Disc method

Learning goals

Identifies the axis and cross-sectional geometry.
Writes the correct cross-sectional area formula.
Sets up the volume integral with correct limits.
Distinguishes discs from washers when appropriate.
Evaluates and interprets the resulting volume.

Difficulty: 1 – 5

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Disc method 1

The shaded region shown is revolved about the x-axis. Write and evaluate a disc-method integral for the volume.

Hints

- Read the constant radius and the x-interval from the graph. - Square the radius to obtain each disc's area. - Accumulate the constant disc area across the displayed interval.

Solution

1. From the graph, every disc has radius \(2\), and the region extends from \(x=0\) to \(x=5\). 2. Therefore, \(V=\pi\int_0^5 2^2\,\mathrm{d}x=20\pi\) cubic units.

Answer

\(20\pi\) cubic units

Disc method 2

The shaded region is revolved about the dashed horizontal line shown. At \(x=2\), what is the radius of the resulting disc cross section?

Hints

- Read the y-coordinate of the upper boundary at the specified x-value. - Read the y-coordinate of the dashed axis. - Radius is a distance from the axis of rotation, not an absolute y-value.

Solution

1. At \(x=2\), the upper boundary is at \(y=4\), while the axis of rotation is at \(y=1\). 2. The disc radius is the perpendicular distance \(4-1=3\) units.

Answer

\(3\) units

Disc method 3

Rotate the shaded region about the x-axis. At \(x=1\), is a slice perpendicular to that axis a solid disc or a hollow washer? What is the area of that slice?

Hints

- Does the original region touch the axis of rotation? - The disc's radius is measured perpendicular to the rotation axis. - Use the area of a circle, not its circumference.

Solution

1. The shaded region reaches the x-axis, so its rotated cross section is a solid disc, with no hole. 2. The radius at \(x=1\) is the height \(y=x=1\). 3. Its area is \(\pi r^2=\pi\) square units.

Answer

A solid disc of area \(\pi\) square units.

Disc method 4

Rotate the shaded region about the dashed line. Find the radius and area of a disc cross section.

Hints

- Measure radius from the rotation axis, not from the x-axis. - The distance between two horizontal lines is the difference of their y-values.

Solution

1. The axis of rotation is the lower boundary \(y=2\). 2. The radius reaches from \(y=2\) to \(y=4\), so \(r=4-2=2\). 3. Each cross section has area \(\pi r^2=4\pi\) square units.

Answer

Radius: \(2\). Disc area: \(4\pi\) square units.

Disc method 5

A solid cylinder is cut by a plane parallel to its circular bases. What shape is the cross section, and what is its area?

Hints

- A cross section parallel to a cylinder's base has the base's shape. - Use the base radius to find its area.

Solution

1. A slice parallel to the cylinder's base is a circle with the same radius, \(3\,\text{cm}\). 2. Its area is \(\pi\cdot3^2=9\pi\,\text{cm}^2\).

Answer

A circular disc with area \(9\pi\,\text{cm}^2\).

Disc method 6

Find the volume of the solid formed by rotating the region under \(f(x)=\sqrt{3x+1}\) over \([1,5]\) about the x-axis.

Hints

- Recall the disc-method formula for rotation about the x-axis. - Simplify the squared square root. - Find an antiderivative of the resulting linear function. - Include the factor \(\pi\).

Solution

1. By the disc method, \(V=\pi\int_1^5(\sqrt{3x+1})^2\,\text{d}x=\pi\int_1^5(3x+1)\,\text{d}x\). 2. \(V=\pi[\frac{3}{2}x^2+x]_1^5=\pi(42.5-2.5)=40\pi\).

Answer

\(40\pi\) cubic units

Disc method 7

For \(f(x)=\frac{2}{x}\), the region under the graph from \(x=1\) to \(x=4\) is rotated about the x-axis. Find the exact volume.

Hints

- What expression gives the disc radius? - Square the complete reciprocal expression before integrating. - Rewrite the integrand using a negative exponent.

Solution

1. The disc radius is \(\frac{2}{x}\), so \(V=\pi\int_1^4\left(\frac{2}{x}\right)^2\,\mathrm{d}x=4\pi\int_1^4x^{-2}\,\mathrm{d}x\). 2. Therefore, \(V=4\pi\left[-\frac{1}{x}\right]_1^4=3\pi\) cubic units.

Answer

\(3\pi\) cubic units

Disc method 8

The region under \(f(x)=\sqrt{x}\) from \(x=1\) to \(x=4\) is rotated about the x-axis. Find the volume.

Hints

- Use \(f(x)\) as the disc radius. - What happens when you square \(\sqrt{x}\)? - Use the stated endpoints as the integration limits.

Solution

1. The disc radius is \(\sqrt{x}\), so \(V=\pi\int_1^4(\sqrt{x})^2\,\mathrm{d}x=\pi\int_1^4x\,\mathrm{d}x\). 2. Therefore, \(V=\pi\left[\frac{x^2}{2}\right]_1^4=\frac{15\pi}{2}\) cubic units.

Answer

\(\frac{15\pi}{2}\) cubic units

Disc method 9

The region under \(f(x)=\sqrt{x^2+5}\) from \(x=0\) to \(x=2\) is rotated about the x-axis. Find the volume.

Hints

- Use the function value as the disc radius. - Squaring the radius removes the square root. - Integrate the resulting polynomial term by term.

Solution

1. The disc radius is \(\sqrt{x^2+5}\), so \(V=\pi\int_0^2(\sqrt{x^2+5})^2\,\mathrm{d}x=\pi\int_0^2(x^2+5)\,\mathrm{d}x\). 2. Therefore, \(V=\pi\left[\frac{x^3}{3}+5x\right]_0^2=\frac{38\pi}{3}\) cubic units.

Answer

\(\frac{38\pi}{3}\) cubic units

Disc method 10

A region is revolved about the y-axis, producing disc cross sections of radius \(R(y)\) for \(0\le y\le5\). The average value of \([R(y)]^2\) on this interval is \(7\) square units. Find the volume of the solid.

Hints

- Translate the stated average into an accumulated quantity over the interval. - Identify which squared measurement determines each disc's area. - Keep the factor common to every circular cross section until the end.

Solution

1. The average-value statement gives \(\frac15\int_0^5[R(y)]^2\,\mathrm{d}y=7\). 2. Therefore, \(\int_0^5[R(y)]^2\,\mathrm{d}y=35\). 3. The disc-method volume is \(V=\pi\int_0^5[R(y)]^2\,\mathrm{d}y=35\pi\).

Answer

\(35\pi\) cubic units

Disc method 11

The region between \(y=6-x\) and the x-axis for \(0\le x\le6\) is revolved about the x-axis. Bence writes \(V=\pi\int_0^6\left(\frac{6-x}{2}\right)^2\,\mathrm{d}x\). Explain the error and give the correct volume.

Hints

- Is \(6-x\) the full distance across a disc or the distance from the axis to one boundary? - A radius is measured from the axis of rotation to the generating curve. - Correct the radius before evaluating the integral.

Solution

1. The vertical distance \(6-x\) runs from the axis of rotation to the boundary, so it is already the disc radius, not a diameter. 2. The correct integral is \(V=\pi\int_0^6(6-x)^2\,\mathrm{d}x\). 3. Evaluating gives \(V=72\pi\) cubic units.

Answer

The radius was incorrectly halved. The correct volume is \(72\pi\) cubic units.

Disc method 12

A nonnegative function \(f\) on \([a,b]\) generates a solid of volume \(12\pi\) when the region under its graph is revolved about the x-axis. What volume is generated by the region under \(g(x)=3f(x)\) on the same interval?

Hints

- Track how a vertical scale changes every disc radius. - Volume depends on the square of the radius scale.

Solution

1. The original volume is \(\pi\int_a^b[f(x)]^2\,\mathrm{d}x=12\pi\). 2. For \(g(x)=3f(x)\), the squared radius is \(9[f(x)]^2\). 3. Therefore, the new volume is \(9(12\pi)=108\pi\).

Answer

\(108\pi\) cubic units

Disc method 13

The shaded region is revolved about the dashed horizontal line shown. For slices perpendicular to the x-axis, decide whether the cross sections are discs or washers. Then write the radius function and the volume integral. Do not evaluate the integral.

Hints

- Does each shaded vertical slice touch the axis of rotation? - A hole appears only when the slice stays a positive distance from the axis. - Measure the radius as the vertical distance from the dashed axis to the upper boundary.

Solution

1. The lower edge of the region lies on the axis of rotation, so each perpendicular slice has no hole and produces a disc. 2. The upper boundary is \(y=4-x\) and the axis is \(y=1\), so the radius is \(R(x)=(4-x)-1=3-x\). 3. The volume setup is \(V=\pi\int_0^3(3-x)^2\,\mathrm{d}x\).

Answer

Disc cross sections; \(R(x)=3-x\); \(V=\pi\int_0^3(3-x)^2\,\mathrm{d}x\).

Disc method 14

Rotate the shaded region about the y-axis. Use discs perpendicular to the y-axis to find the solid's volume.

Hints

- A cross section perpendicular to the y-axis is horizontal in the original region. - Express the right-hand boundary as \(x\) in terms of \(y\). - Square the radius before integrating over height.

Solution

1. A horizontal slice at height \(y\) reaches from \(x=0\) to \(x=\sqrt{y}\); revolving it gives a disc of radius \(\sqrt{y}\). 2. Its area is \(\pi(\sqrt{y})^2=\pi y\). 3. The height goes from \(0\) to \(4\), so \(V=\pi\int_0^4y\,\mathrm{d}y=8\pi\) cubic units.

Answer

\(V=\pi\int_0^4y\,\mathrm{d}y=8\pi\) cubic units.

Disc method 15

Rotating the region under a positive function \(f(x)\) on \([0,2]\) about the x-axis gives the disc-method volume \(V=\pi\int_0^2(3-x)^2\,\mathrm{d}x\). What is the function \(f(x)\), and what is the disc radius at \(x=1\)?

Hints

- Compare the volume formula to \(\pi\int_a^b r(x)^2\,\mathrm{d}x\). - The positivity condition selects the appropriate square root. - Evaluate the radius at the requested location.

Solution

1. In the disc method, the squared expression is the radius squared. 2. Because \(f\) is positive and its region reaches the axis, \(f(x)=3-x\). 3. At \(x=1\), the radius is \(f(1)=2\).

Answer

\(f(x)=3-x\). Radius at \(x=1\): \(2\).

Disc method 16

Two solids are formed by rotating regions under positive graphs about the x-axis, both from \(x=0\) to \(x=1\). The second graph is twice as high as the first at every \(x\). How many times as large is the second solid's volume? Explain without evaluating an integral.

Hints

- Volume adds the areas of discs with matching thicknesses. - How does the area of a circle change when its radius doubles? - The two solids have the same interval of integration.

Solution

1. At each \(x\), the second disc radius is \(2r(x)\) rather than \(r(x)\). 2. Its cross-sectional area is \(\pi(2r(x))^2=4\pi r(x)^2\). 3. Every slice has four times the area, so the total volume is four times as large.

Answer

\(4\) times as large, because doubling each disc radius multiplies its area by \(4\).

Disc method 17

Rotate the shaded region about the dashed line. Find the volume of the resulting solid using discs.

Hints

- Measure each radius from the dashed line to the sloping edge, not from the x-axis. - Express that distance in terms of \(x\). - A disc's area is based on the square of its radius.

Solution

1. The dashed line is \(y=2\). At position \(x\), the region extends to \(y=x+3\), so the disc radius is \(r(x)=(x+3)-2=x+1\). 2. The disc area is \(\pi(x+1)^2\). The region runs from \(x=0\) to \(x=2\). 3. Thus \(V=\pi\int_0^2(x+1)^2\,\mathrm{d}x=\pi\left[\frac{(x+1)^3}{3}\right]_0^2=\frac{26\pi}{3}\) cubic units.

Answer

\(V=\pi\int_0^2(x+1)^2\,\mathrm{d}x=\frac{26\pi}{3}\) cubic units.

Disc method 18

The region between the line \(f(x)=\frac{x}{2}\) and the x-axis on \([k,3k]\), where \(k>0\), is revolved about the x-axis. Using the disc method, set up and evaluate the volume integral. Identify the solid and check using the frustum formula \(V=\frac{\pi h}{3}(R^2+Rr+r^2)\).

Hints

- For rotation about the x-axis, what vertical distance gives the disc radius? - The required answer starts with a disc-method integral, not the frustum formula. - After evaluating the integral, identify the two end radii and the frustum height for the check.

Solution

1. The disc radius is \(R(x)=\frac{x}{2}\), so \(V=\pi\int_k^{3k}\left(\frac{x}{2}\right)^2\,\text{d}x\). 2. Evaluating gives \(V=\frac{\pi}{12}[(3k)^3-k^3]=\frac{13\pi}{6}k^3\). 3. The solid is a conical frustum with height \(2k\) and radii \(\frac{k}{2}\) and \(\frac{3k}{2}\). The formula \(V=\frac{\pi h}{3}(R^2+Rr+r^2)\) also gives \(\frac{13\pi}{6}k^3\).

Answer

Disc-method setup: \(V=\pi\int_k^{3k}\left(\frac{x}{2}\right)^2\,\text{d}x\) Volume: \(\frac{13\pi}{6}k^3\) cubic units The solid is a conical frustum, and its geometric formula gives the same volume.

Disc method 19

The region under \(f(x)=\sqrt{rx}\) and above the x-axis, where \(r>0\), is revolved about the x-axis over \([0,r]\). Find the resulting solid's volume in terms of \(r\). Then compare it with the volume of a cylinder that has the same height \(r\) and the same maximum radius as the solid.

Hints

- Recall the disc-method formula for rotation about the x-axis. - What happens when \(\sqrt{rx}\) is squared? - Where is the radius largest on \([0,r]\)? - Use the cylinder-volume formula for the comparison.

Solution

1. By the disc method, \(V=\pi\int_0^r(\sqrt{rx})^2\,\text{d}x=\pi\int_0^r rx\,\text{d}x\). 2. Thus \(V=\pi r[\frac12x^2]_0^r=\frac12\pi r^3\). 3. The maximum radius occurs at \(x=r\), where \(f(r)=r\). The comparison cylinder has radius \(r\) and height \(r\), so its volume is \(\pi r^3\). 4. Therefore, the solid of revolution has half the cylinder's volume.

Answer

\(V=\frac12\pi r^3\), which is half the cylinder's volume \(\pi r^3\).

Disc method 20

A solid of revolution has volume \(V=\pi\int_0^3(x+2)^2\,\text{d}x\). 1. Identify the boundary function \(f\) that is rotated, and describe the geometric shape of the solid. 2. Evaluate the volume and express the result as a multiple of \(\pi\). 3. Find the radii of the two circular bases.

Hints

- Identify the function that is squared in the volume integral. - What solid forms when a slanted line segment is rotated about an axis? - Use the power rule after treating \(x+2\) as a single expression. - The endpoint radii are the function values at the integration limits.

Solution

1. Comparing with \(V=\pi\int_a^b[f(x)]^2\,\text{d}x\) gives \(f(x)=x+2\). Rotating this line segment about the x-axis produces a right conical frustum. 2. \(V=\pi[\frac{1}{3}(x+2)^3]_0^3=\pi\left(\frac{125}{3}-\frac{8}{3}\right)=39\pi\). 3. The endpoint radii are \(f(0)=2\) and \(f(3)=5\).

Answer

1. \(f(x)=x+2\); a right conical frustum 2. \(39\pi\) 3. \(2\) and \(5\)

Disc method 21

Revolving the region between \(h(x)=\sqrt{x}\) and the x-axis on \([0,a]\) about the x-axis creates a paraboloid of revolution. 1. Find \(a>0\) so that the volume is exactly \(32\pi\) cubic units. 2. A cone is formed by revolving the region between \(k(x)=\frac{x}{2}\) and the x-axis over the same interval. Find its volume using the value of \(a\) from part 1. 3. Find the ratio \(V_{\text{paraboloid}}:V_{\text{cone}}\).

Hints

- First write the paraboloid's volume as a function of \(a\). - Simplify \((\sqrt{x})^2\). - When squaring \(\frac12x\), square both the coefficient and the variable. - Simplify the ratio by canceling common factors.

Solution

1. \(V_{\text{paraboloid}}=\pi\int_0^a(\sqrt{x})^2\,\text{d}x=\frac12\pi a^2\). Setting this equal to \(32\pi\) gives \(a=8\). 2. \(V_{\text{cone}}=\pi\int_0^8\left(\frac{x}{2}\right)^2\,\text{d}x=\frac{\pi}{4}[\frac{x^3}{3}]_0^8=\frac{128\pi}{3}\). 3. The ratio is \(32\pi:\frac{128\pi}{3}=3:4\).

Answer

1. \(a=8\) 2. \(V_{\text{cone}}=\frac{128\pi}{3}\) cubic units 3. \(3:4\)

Disc method 22

A goblet-shaped container has circular horizontal cross sections of radius \(f(x)=\sqrt{3x}\) for \(0\le x\le10\), where \(x\) is height above the bottom and both measurements are in centimeters. The container holds \(150\,\text{mL}\) of liquid. Find the liquid height exactly and to the nearest hundredth of a centimeter.

Hints

- Use the liquid height as the upper limit of a disc-method volume integral. - How are milliliters related to cubic centimeters? - After integrating, solve the resulting equation for the positive height.

Solution

1. If the liquid reaches height \(h\), its volume is \(V(h)=\pi\int_0^h[f(x)]^2\,\text{d}x\). 2. Thus \(V(h)=\pi\int_0^h3x\,\text{d}x=\frac32\pi h^2\). 3. Since \(150\,\text{mL}=150\,\text{cm}^3\), solve \(\frac32\pi h^2=150\). 4. Therefore, \(h=\sqrt{\frac{100}{\pi}}=\frac{10}{\sqrt{\pi}}\,\text{cm}\approx5.64\,\text{cm}\).

Answer

\(\frac{10}{\sqrt{\pi}}\,\text{cm}\approx5.64\,\text{cm}\)

Disc method 23

A bowl has circular horizontal cross sections of radius \(f(x)=e^{0.2x}\) at height \(x\), for \(0\le x\le8\), with all lengths in centimeters. The bowl contains \(60\,\text{mL}\) of water. Find the water depth exactly and to the nearest hundredth of a centimeter.

Hints

- Square the exponential radius before integrating. - Use the water depth as the variable upper limit. - After isolating the exponential expression, what inverse operation solves for \(h\)?

Solution

1. If the water reaches height \(h\), its volume is \(V(h)=\pi\int_0^h(e^{0.2x})^2\,\text{d}x=\pi\int_0^h e^{0.4x}\,\text{d}x\). 2. Thus \(V(h)=2.5\pi(e^{0.4h}-1)\). 3. Since \(60\,\text{mL}=60\,\text{cm}^3\), solve \(2.5\pi(e^{0.4h}-1)=60\). 4. Then \(e^{0.4h}=1+\frac{24}{\pi}\), so \(h=\frac52\ln\left(1+\frac{24}{\pi}\right)\,\text{cm}\approx5.39\,\text{cm}\), within the modeled height.

Answer

\(\frac52\ln\left(1+\frac{24}{\pi}\right)\,\text{cm}\approx5.39\,\text{cm}\)

Disc method 24

The region bounded by \(f(x)=x\sqrt{6-x}\) and the x-axis on \([0,6]\) is rotated about the x-axis. Find the volume of the resulting solid.

Hints

- What function gives the disc radius at position \(x\)? - Square the entire radius function before integrating. - Simplify the squared radical to a polynomial.

Solution

1. A representative disc has radius \(f(x)=x\sqrt{6-x}\), so \(V=\pi\int_0^6[f(x)]^2\,\mathrm{d}x\). 2. \((x\sqrt{6-x})^2=x^2(6-x)=6x^2-x^3\). 3. Therefore, \(V=\pi\left[2x^3-\frac{1}{4}x^4\right]_0^6=108\pi\) cubic units.

Answer

\(108\pi\) cubic units

Disc method 25

The region bounded by \(g(x)=\frac{3}{x+1}\), the x-axis, the y-axis, and the line \(x=2\) is rotated about the x-axis. Find the volume of the resulting solid.

Hints

- Identify the left and right boundaries of the region. - Square the entire function before integrating. - Rewrite the reciprocal square using a negative exponent. - Apply the power rule with the linear expression \(x+1\).

Solution

1. The bounds are \(x=0\) and \(x=2\). 2. By the disc method, \(V=\pi\int_0^2\left(\frac{3}{x+1}\right)^2\,\mathrm{d}x=9\pi\int_0^2(x+1)^{-2}\,\mathrm{d}x\). 3. \(V=\pi[-\frac{9}{x+1}]_0^2=\pi(-3+9)=6\pi\).

Answer

\(6\pi\) cubic units

Disc method 26

The region between \(f(x)=2e^{-0.25x}\) and the x-axis over \([0,4]\) is rotated about the x-axis. Find the volume of the resulting solid.

Hints

- Square both the coefficient and the exponential factor. - Integrate the exponential function with its linear exponent. - Check signs and use \(e^0=1\) when evaluating the limits. - Give an exact value before rounding.

Solution

1. By the disc method, \(V=\pi\int_0^4(2e^{-0.25x})^2\,\text{d}x=4\pi\int_0^4e^{-0.5x}\,\text{d}x\). 2. \(V=4\pi[-2e^{-0.5x}]_0^4=8\pi(1-e^{-2})\).

Answer

\(V=8\pi(1-e^{-2})\) cubic units, approximately \(21.73\) cubic units

Disc method 27

Derive the volume formula for a sphere of radius \(R\). Rotate the region between \(y=\sqrt{R^2-x^2}\) and the x-axis, for \(-R\le x\le R\), about the x-axis. Show that \(V=\frac{4}{3}\pi R^3\).

Hints

- What shape does a thin vertical strip make after rotation? - Which distance in the original region determines the area of a cross section? - Can the symmetry of the two halves reduce the work?

Solution

1. Rotating the region produces discs with radius \(\sqrt{R^2-x^2}\). 2. Therefore, \(V=\pi\int_{-R}^R(R^2-x^2)\,\text{d}x\). 3. Evaluating gives \(V=\pi[R^2x-\frac{x^3}{3}]_{-R}^R=\frac43\pi R^3\).

Answer

\(V=\frac43\pi R^3\)

Disc method 28

Let \(f(x)=\frac{2}{\sqrt[4]{x^3}}\) for \(x>0\). The graph and the x-axis bound two unbounded regions: one over \([1,\infty)\) and one over \((0,1]\). Determine whether the solids formed by rotating these regions about the x-axis have finite volume. Find each finite volume.

Hints

- Rewrite the fourth root using a rational exponent. - Use the disc-method volume formula. - Treat both integrals as improper. - Compare the behavior of reciprocal powers as \(x\to\infty\) and as \(x\to0^+\).

Solution

1. \((f(x))^2=\frac{4}{x^{3/2}}=4x^{-3/2}\), and an antiderivative is \(-8x^{-1/2}\). 2. Over \([1,\infty)\), \(V_1=\pi\lim_{b\to\infty}[-8x^{-1/2}]_1^b=\pi(0+8)=8\pi\). This volume is finite. 3. Over \((0,1]\), \(V_2=\pi\lim_{a\to0^+}[-8x^{-1/2}]_a^1\). Since \(8a^{-1/2}\to\infty\), the volume diverges.

Answer

Over \([1,\infty)\), the volume is \(8\pi\) cubic units. Over \((0,1]\), the volume is infinite.

Disc method 29

A wooden display pedestal is modeled by revolving the region under \(f(x)=\sqrt{0.2x^2+4}\) over \([-4,4]\) about the x-axis. One model unit represents \(10\,\text{cm}\) in the actual pedestal. a) Find its volume in liters, exactly and to the nearest hundredth of a liter. b) A cylindrical hole of radius \(1\) model unit extends \(2\) model units into the end at \(x=4\), along the x-axis. Find the remaining volume in liters, exactly and to the nearest hundredth.

Hints

- Square the radius function before integrating with the disc method. - How does a linear scale factor of \(10\) affect cubic units? - For the recess, subtract the volume of a cylinder from the original solid.

Solution

1. By the disc method, \(V=\pi\int_{-4}^{4}(0.2x^2+4)\,\text{d}x=\frac{608\pi}{15}\) model cubic units. 2. One model cubic unit is \(1000\,\text{cm}^3=1\,\text{L}\). Thus the pedestal volume is \(\frac{608\pi}{15}\,\text{L}\approx127.34\,\text{L}\). 3. The recess has volume \(\pi(1)^2(2)=2\pi\) model cubic units, or \(2\pi\,\text{L}\). 4. The remaining volume is \(\frac{608\pi}{15}-2\pi=\frac{578\pi}{15}\,\text{L}\approx121.06\,\text{L}\).

Answer

a) \(\frac{608\pi}{15}\,\text{L}\approx127.34\,\text{L}\) b) \(\frac{578\pi}{15}\,\text{L}\approx121.06\,\text{L}\)

Disc method 30

A container shaped like a conical frustum has interior radius \(f(x)=0.2x+3\) at height \(x\), for \(0\le x\le10\). All dimensions are in centimeters, and the bottom is at \(x=0\). Its capacity is the volume of the circular cross sections stacked from the bottom. a) Find the volume of liquid when filled to a height of \(8\,\text{cm}\). Give an exact value and a value rounded to the nearest hundredth of a cubic centimeter. b) Write an equation without an integral sign that could be used to find the liquid height \(h\) when the container holds exactly \(300\,\text{cm}^3\) of liquid.

Hints

- Use the fill height as the upper limit of the disc-method integral. - Evaluate the same antiderivative with a variable upper limit \(h\). - What equation makes the resulting volume equal to \(300\,\text{cm}^3\)?

Solution

1. At height \(8\), \(V(8)=\pi\int_0^8(0.2x+3)^2\,\text{d}x\). 2. An antiderivative is \(\frac{x^3}{75}+0.6x^2+9x\), so \(V(8)=\frac{8792\pi}{75}\,\text{cm}^3\approx368.28\,\text{cm}^3\). 3. Replacing the upper limit by \(h\) gives \(V(h)=\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)\). 4. Therefore, the required equation is \(\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)=300\).

Answer

a) \(\frac{8792\pi}{75}\,\text{cm}^3\approx368.28\,\text{cm}^3\) b) \(\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)=300\)

Disc method 31

A bowl's interior radius is \(g(x)=\sqrt{x+4}\) at height \(x\) for \(0\le x\le12\). Its interior has circular cross sections of this radius; all dimensions are in centimeters. a) Find the bowl's maximum capacity exactly and to the nearest hundredth of a cubic centimeter. b) The bowl contains \(150\,\text{cm}^3\) of water. Find the water depth exactly and to the nearest hundredth of a centimeter.

Hints

- Squaring the interior radius \(g(x)\) simplifies the disc-area integrand. - For part b), use the water depth as a variable upper limit. - After integrating, solve the resulting quadratic equation and keep the physically meaningful root.

Solution

1. The full capacity is \(V_{\max}=\pi\int_0^{12}(x+4)\,\text{d}x=120\pi\,\text{cm}^3\approx376.99\,\text{cm}^3\). 2. For water depth \(h\), \(\pi\int_0^h(x+4)\,\text{d}x=150\), so \(\pi\left(\frac{h^2}{2}+4h\right)=150\). 3. Solving the quadratic and taking the positive root gives \(h=-4+\sqrt{16+\frac{300}{\pi}}\,\text{cm}\approx6.56\,\text{cm}\).

Answer

a) \(120\pi\,\text{cm}^3\approx376.99\,\text{cm}^3\) b) \(-4+\sqrt{16+\frac{300}{\pi}}\,\text{cm}\approx6.56\,\text{cm}\)

Disc method 32

A right circular cone has base radius \(r\) and height \(h\). a) Find the equation of a line through the origin, \(f(x)=mx\), so that revolving the region between the line and the x-axis over \([0,h]\) produces the cone. b) Use integration to derive the cone-volume formula \(V=\frac13\pi r^2h\).

Hints

- Determine the slope from the two endpoint radii. - Use the line as the disc radius function. - Treat \(r\) and \(h\) as constants while integrating with respect to \(x\).

Solution

1. The line passes through \((0,0)\) and \((h,r)\), so its slope is \(m=\frac{r}{h}\). Thus \(f(x)=\frac{r}{h}x\). 2. By the disc method, \(V=\pi\int_0^h\left(\frac{r}{h}x\right)^2\,\text{d}x\). 3. Therefore \(V=\pi\frac{r^2}{h^2}[\frac{x^3}{3}]_0^h=\frac13\pi r^2h\).

Answer

a) \(f(x)=\frac{r}{h}x\) b) \(V=\frac13\pi r^2h\)

Disc method 33

The inside profile of a designer goblet is modeled by \(y=ax^3\), where \(x\) is the inside radius and \(y\) is the height above the bottom, both in centimeters. At a fill height of \(8\,\text{cm}\), the radius of the water surface is \(4\,\text{cm}\). a) Find \(a\). b) Find the capacity of the goblet up to a height of \(8\,\text{cm}\).

Hints

- Use the given radius and height as a point on the profile. - Solve the profile equation for radius \(x\) in terms of height \(y\). - A horizontal slice rotates into a disc whose radius is \(x(y)\).

Solution

1. The point \((4,8)\) lies on the profile, so \(8=a(4)^3\), giving \(a=\frac{1}{8}\). 2. Solve for radius in terms of height: \(x=(8y)^{1/3}\), so \(x^2=4y^{2/3}\). 3. Horizontal slices rotate about the y-axis into discs, so \(V=\pi\int_0^8x(y)^2\,\mathrm{d}y=4\pi\int_0^8y^{2/3}\,\mathrm{d}y\). 4. Therefore, \(V=\frac{384\pi}{5}\,\text{cm}^3\approx241.27\,\text{cm}^3\).

Answer

a) \(a=\frac{1}{8}\) b) \(\frac{384\pi}{5}\,\text{cm}^3\approx241.27\,\text{cm}^3\)

Disc method 34

A rotationally symmetric component is formed by rotating the region under \(f(x)=\frac{3}{\sqrt{2x+1}}\) over \([0,b]\) about the x-axis. a) Derive a formula for the volume \(V(b)\). b) Find the value of \(b\) for which the volume is exactly \(9\pi\) cubic units.

Hints

- Substitute the radius function into the disc-method formula. - Simplify the squared radical before integrating. - Use the logarithmic antiderivative for a reciprocal linear expression. - Apply the exponential function to undo the logarithm.

Solution

1. By the disc method, \(V(b)=\pi\int_0^b\left(\frac{3}{\sqrt{2x+1}}\right)^2\,\mathrm{d}x=9\pi\int_0^b\frac{1}{2x+1}\,\mathrm{d}x\). 2. Therefore, \(V(b)=\frac{9\pi}{2}[\ln(2x+1)]_0^b=\frac{9\pi}{2}\ln(2b+1)\). 3. Set the formula equal to \(9\pi\): \(\frac{9\pi}{2}\ln(2b+1)=9\pi\). 4. Then \(\ln(2b+1)=2\), so \(2b+1=e^2\) and \(b=\frac{e^2-1}{2}\approx3.19\).

Answer

a) \(V(b)=\frac{9\pi}{2}\ln(2b+1)\) b) \(b=\frac{e^2-1}{2}\approx3.19\)

Disc method 35

The region bounded by \(f(x)=(x+1)\sqrt{2-x}\) and the x-axis is rotated about the x-axis. Find the volume of the resulting solid.

Hints

- Use the zeros of the function to determine the interval. - Apply the disc-method formula. - Expand the square of the function before integrating. - Be careful with signs when evaluating at the lower limit.

Solution

1. The zeros are \(x=-1\) and \(x=2\), so these are the limits of integration. 2. By the disc method, \(V=\pi\int_{-1}^{2}(f(x))^2\,\mathrm{d}x\). 3. \((f(x))^2=(x+1)^2(2-x)=-x^3+3x+2\). 4. Therefore, \(V=\pi\left[-\frac{x^4}{4}+\frac{3x^2}{2}+2x\right]_{-1}^{2}\). 5. The definite integral is \(6-(-\frac{3}{4})=\frac{27}{4}\), so \(V=\frac{27\pi}{4}\approx21.21\) cubic units.

Answer

\(\frac{27\pi}{4}\) cubic units, or approximately \(21.21\) cubic units

Disc method 36

The region between \(f(x)=\sqrt{x\sin x}\) and the x-axis over \([0,\pi]\) is revolved about the x-axis. Find the exact volume of the resulting solid.

Hints

- Squaring the radius function removes the square root. - Think about how to integrate the product \(x\sin x\). - Evaluate the sine and cosine terms at both endpoints. - Include the factor \(\pi\) from the disc area.

Solution

1. By the disc method, \(V=\pi\int_0^\pi[f(x)]^2\,\text{d}x=\pi\int_0^\pi x\sin x\,\text{d}x\). 2. Integration by parts gives \(\int x\sin x\,\text{d}x=-x\cos x+\sin x\). 3. Thus \(\int_0^\pi x\sin x\,\text{d}x=[-x\cos x+\sin x]_0^\pi=\pi\). 4. Therefore, \(V=\pi^2\) cubic units.

Answer

\(\pi^2\) cubic units

Disc method 37

A designer models a wooden paperweight as the solid formed by revolving the region between \(f(x)=2\sqrt{x}\) and the x-axis over \([0,5]\) about the x-axis. One coordinate unit represents \(1\,\text{cm}\). 1. Find the volume of the paperweight. 2. The wood has density \(0.8\,\text{g/cm}^3\). Find the mass of the paperweight.

Hints

- Use the disc-method formula for the filled region. - Simplify the squared square root before integrating. - Use mass equals density times volume. - Keep the units consistent.

Solution

1. By the disc method, \(V=\pi\int_0^5(2\sqrt{x})^2\,\text{d}x=\pi\int_0^5 4x\,\text{d}x\). 2. \(V=\pi[2x^2]_0^5=50\pi\,\text{cm}^3\approx157.08\,\text{cm}^3\). 3. Using \(m=\rho V\), \(m=0.8(50\pi)=40\pi\,\text{g}\approx125.66\,\text{g}\).

Answer

1. \(50\pi\,\text{cm}^3\approx157.08\,\text{cm}^3\) 2. \(40\pi\,\text{g}\approx125.66\,\text{g}\)

Disc method 38

The region between \(f(x)=\sqrt{x}e^x\) and the x-axis over \([0,1]\) is revolved about the x-axis. Find the exact volume of the resulting solid.

Hints

- Use the disc-method formula for the region. - Square both factors in the radius function. - Consider integration by parts for \(xe^{2x}\). - Include both terms of the antiderivative when evaluating the lower limit.

Solution

1. By the disc method, \(V=\pi\int_0^1[f(x)]^2\,\text{d}x=\pi\int_0^1xe^{2x}\,\text{d}x\). 2. Integration by parts gives \(\int xe^{2x}\,\text{d}x=\frac12xe^{2x}-\frac14e^{2x}\). 3. Therefore \(V=\pi[\frac12xe^{2x}-\frac14e^{2x}]_0^1\). 4. Evaluating gives \(V=\frac{\pi(e^2+1)}{4}\).

Answer

\(\frac{\pi(e^2+1)}{4}\) cubic units

Disc method 39

The graphs of \(f(x)=2-e^x\) and \(g(x)=2-e^{-x}\), together with the x-axis, enclose a symmetric region. Find the volume generated when the region is rotated about the x-axis.

Hints

- Find the x-intercept of each exponential boundary. - How does the relation \(f(x)=g(-x)\) simplify the volume integral? - Expand the squared radius before integrating.

Solution

1. The x-intercepts are \(x=\ln2\) for \(f\) and \(x=-\ln2\) for \(g\), and the graphs meet at \((0,1)\). 2. Since \(f(x)=g(-x)\), the region is symmetric about the y-axis. Thus \(V=2\pi\int_0^{\ln2}(2-e^x)^2\,\mathrm{d}x\). 3. Expand: \((2-e^x)^2=4-4e^x+e^{2x}\), with antiderivative \(4x-4e^x+\frac{1}{2}e^{2x}\). 4. Evaluation gives \(V=\pi(8\ln2-5)\) cubic units.

Answer

\(\pi(8\ln2-5)\) cubic units

Disc method 40

The graphs of \(f(x)=\sin x\) and \(g(x)=\cos x\), together with the x-axis, enclose a region on \([0,\frac{\pi}{2}]\). Find the volume generated when the region is rotated about the x-axis.

Hints

- Find where the sine and cosine graphs intersect. - Identify the upper boundary on each part of the interval. - Use the symmetry of sine and cosine on this interval. - Apply a power-reduction identity to integrate the squared trigonometric function.

Solution

1. The graphs intersect where \(\sin x=\cos x\), so \(x=\frac{\pi}{4}\). 2. The upper boundary is \(\sin x\) on \([0,\frac{\pi}{4}]\) and \(\cos x\) on \([\frac{\pi}{4},\frac{\pi}{2}]\). 3. By symmetry, \(V=2\pi\int_0^{\pi/4}\sin^2x\,\mathrm{d}x\). 4. Using \(\sin^2x=\frac{1}{2}(1-\cos2x)\), \(V=\pi\left[x-\frac{1}{2}\sin2x\right]_0^{\pi/4}\). 5. Therefore, \(V=\frac{\pi^2}{4}-\frac{\pi}{2}\approx0.90\) cubic units.

Answer

\(\frac{\pi^2}{4}-\frac{\pi}{2}\) cubic units, or approximately \(0.90\) cubic units

Disc method 41

Let \(g(x)=(4-x)\sqrt{x}\) on \([0,4]\). The region between the graph and the x-axis is rotated about the x-axis. a) Find where the solid has its maximum radius and find that radius. b) Find the volume of the solid.

Hints

- Differentiate the radius function to locate an interior maximum. - Check how the derivative sign changes around the critical point. - Square the entire radius function before setting up the disc-method integral.

Solution

1. For \(x>0\), \(g'(x)=\frac{4-3x}{2\sqrt{x}}\). 2. The derivative changes from positive to negative at \(x=\frac{4}{3}\), so the maximum radius is \(g\left(\frac{4}{3}\right)=\frac{16\sqrt3}{9}\). 3. By the disc method, \(V=\pi\int_0^4[(4-x)\sqrt{x}]^2\,\mathrm{d}x=\pi\int_0^4(16x-8x^2+x^3)\,\mathrm{d}x\). 4. Therefore, \(V=\frac{64\pi}{3}\) cubic units.

Answer

a) The maximum radius is \(\frac{16\sqrt3}{9}\), occurring at \(x=\frac{4}{3}\). b) \(\frac{64\pi}{3}\) cubic units

Disc method 42

A horizontal designer vase has interior radius \(f(x)=0.1x^2-x+5\) centimeters for \(0\le x\le8\), where \(x\) is measured in centimeters along the vase's axis. Rotating the radius profile about the x-axis forms the interior. Find the capacity of the vase to the nearest tenth of a cubic centimeter.

Hints

- The given profile is the radius of each disc perpendicular to the x-axis. - Square the radius before integrating. - Round only after evaluating the definite integral.

Solution

1. By the disc method, \(V=\pi\int_0^8(0.1x^2-x+5)^2\,\mathrm{d}x\). 2. Evaluating the polynomial integral gives \(V=\frac{30776\pi}{375}\,\text{cm}^3\). 3. Therefore, the capacity is \(V\approx257.8\,\text{cm}^3\).

Answer

\(257.8\,\text{cm}^3\)

Disc method 43

A drop-shaped wooden sculpture is formed by rotating the radius profile \(f(x)=0.1(x+2)\sqrt{25-x}\) for \(0\le x\le25\) about the x-axis. One coordinate unit represents \(1\,\text{cm}\). Find the volume exactly and to the nearest tenth of a cubic centimeter.

Hints

- Treat \(f(x)\) as the radius of each perpendicular disc. - Square the complete radius profile before integrating. - Keep the exact value through the integration, then round at the end.

Solution

1. By the disc method, \(V=\pi\int_0^{25}[0.1(x+2)\sqrt{25-x}]^2\,\mathrm{d}x\). 2. Evaluating the resulting polynomial integral gives \(V=\frac{7075\pi}{16}\,\text{cm}^3\). 3. Numerically, \(V\approx1389.2\,\text{cm}^3\).

Answer

\(\frac{7075\pi}{16}\,\text{cm}^3\approx1389.2\,\text{cm}^3\)

Disc method 44

The parabola \(f(x)=4-x^2\) and the x-axis enclose a region. Find the volume generated when this region is rotated about the x-axis.

Hints

- Find the x-intercepts to determine the bounds. - Square the radius function before integrating. - The integrand is even, so symmetry can shorten the computation.

Solution

1. The x-intercepts are \(-2\) and \(2\), so \(V=\pi\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x\). 2. Expand: \((4-x^2)^2=16-8x^2+x^4\). 3. Using symmetry, \(V=2\pi\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^2=\frac{512\pi}{15}\) cubic units.

Answer

\(\frac{512\pi}{15}\) cubic units

Disc method 45

The region under \(f(x)=\sin x\) on \([0,\pi]\) is rotated about the x-axis. Find the volume of the resulting solid.

Hints

- What is the disc radius at position \(x\)? - Which power-reduction identity simplifies \(\sin^2x\)? - Keep the factor \(\pi\) from the disc-area formula.

Solution

1. By the disc method, \(V=\pi\int_0^\pi\sin^2x\,\mathrm{d}x\). 2. Use \(\sin^2x=\frac{1}{2}(1-\cos2x)\). 3. Therefore, \(V=\pi\left[\frac{x}{2}-\frac{\sin2x}{4}\right]_0^\pi=\frac{\pi^2}{2}\) cubic units.

Answer

\(\frac{\pi^2}{2}\) cubic units

Disc method 46

The region between \(y=2+\cos x\) and \(y=2\) for \(0\le x\le\pi\) is revolved about \(y=2\). Find the volume using discs.

Hints

- Measure each radius from the horizontal axis of rotation \(y=2\). - What happens to the sign of \(\cos x\) on the second half of the interval? - Why does squaring the radius allow one integral over the full interval?

Solution

1. The axis of rotation is a boundary of the region, so each perpendicular cross section is a disc. 2. The radius is the distance \(|\cos x|\), and its square is \(\cos^2x\). 3. Therefore, \(V=\pi\int_0^\pi\cos^2x\,\mathrm{d}x=\frac{\pi^2}{2}\) cubic units.

Answer

\(\frac{\pi^2}{2}\) cubic units

Disc method 47

The region bounded by \(y=e^x\), \(x=0\), and \(y=e\) is revolved about the y-axis. Set up and evaluate a disc-method integral.

Hints

- Use slices perpendicular to the vertical axis of rotation. - Rewrite \(y=e^x\) as \(x\) in terms of \(y\). - The horizontal distance from the y-axis to the curve is the disc radius.

Solution

1. Rewrite the curve as \(x=\ln y\) for \(1\le y\le e\). 2. A horizontal slice rotates into a disc with radius \(R(y)=\ln y\). 3. Therefore, \(V=\pi\int_1^e(\ln y)^2\,\mathrm{d}y=\pi(e-2)\) cubic units.

Answer

\(\pi(e-2)\) cubic units

Disc method 48

For \(0\le x\le3\), the region between \(y=kx\) and the x-axis is revolved about the x-axis. Its volume is \(81\pi\). Find the positive value of \(k\).

Hints

- The scale factor in the radius is squared in the volume. - Use the positivity condition after solving the equation.

Solution

1. The disc radius is \(kx\), so \(V=\pi\int_0^3k^2x^2\,\mathrm{d}x=9\pi k^2\). 2. Set \(9\pi k^2=81\pi\). 3. Since \(k>0\), \(k=3\).

Answer

\(k=3\)

Disc method 49

The region bounded by \(x=y(4-y)\), the y-axis, \(y=0\), and \(y=4\) is revolved about the y-axis. Find the volume using discs.

Hints

- Use horizontal slices because the axis of rotation is vertical. - What horizontal distance gives the disc radius at height \(y\)? - Square the entire radius expression before integrating.

Solution

1. A horizontal slice forms a disc with radius \(R(y)=y(4-y)\). 2. Therefore, \(V=\pi\int_0^4y^2(4-y)^2\,\mathrm{d}y\). 3. Expanding and integrating gives \(\int_0^4(16y^2-8y^3+y^4)\,\mathrm{d}y=\frac{512}{15}\). 4. Thus, \(V=\frac{512\pi}{15}\) cubic units.

Answer

\(\frac{512\pi}{15}\) cubic units

Disc method 50

The cone shown has circular cross sections perpendicular to its height. Let \(x\) measure distance from the apex toward the base. Using the dimensions shown, find the radius function \(r(x)\), set up a disc-method integral, and evaluate the cone's volume.

Hints

- The cross-sectional radius changes linearly from the apex to the base. - Use similar triangles to relate radius to distance from the apex. - Disc area is \(\pi r(x)^2\) before integrating along the height.

Solution

1. The radius grows linearly from \(0\) at the apex to \(3\,\text{cm}\) over a height of \(4\,\text{cm}\), so similarity gives \(r(x)=\frac{3}{4}x\). 2. The disc-method setup is \(V=\pi\int_0^4\left(\frac{3x}{4}\right)^2\,\mathrm{d}x\). 3. Evaluating gives \(V=\frac{9\pi}{16}\cdot\frac{64}{3}=12\pi\,\text{cm}^3\).

Answer

\(r(x)=\frac{3}{4}x\), and \(V=12\pi\,\text{cm}^3\).

Disc method 51

Consider \(f_k(x)=x^{-k}\) for \(x\ge1\), where \(k\in\mathbb{R}\). Revolving the region between the graph of \(f_k\) and the x-axis for \(x\ge1\) about the x-axis creates an unbounded solid of revolution. Determine the values of \(k\) for which the solid has finite volume, and give a formula for \(V(k)\).

Hints

- Recall the disc-method formula for a region rotated about the x-axis. - Rewrite the improper integral using a variable upper limit. - When does a negative power of \(x\) approach zero as \(x\to\infty\)? - Treat the logarithmic antiderivative case separately.

Solution

1. The disc method gives \(V=\pi\int_1^\infty[f_k(x)]^2\,\text{d}x=\pi\int_1^\infty x^{-2k}\,\text{d}x\). 2. When \(k\ne\frac12\), an antiderivative is \(\frac{x^{1-2k}}{1-2k}\). The improper integral converges only for \(1-2k<0\), or \(k>\frac12\). 3. For \(k=\frac12\), the integrand is \(x^{-1}\), and the integral diverges logarithmically. 4. For \(k>\frac12\), the integral equals \(\frac{1}{2k-1}\), so \(V(k)=\frac{\pi}{2k-1}\).

Answer

Finite volume for \(k>\frac12\); \(V(k)=\frac{\pi}{2k-1}\) cubic units.

Disc method 52

Let \(g(x)=\frac{1}{\sqrt{x}}\) for \(x>0\). The region between the graph of \(g\), the x-axis, and the lines \(x=0\) and \(x=1\) is unbounded above. Show that the region has finite area. Then determine whether the solid formed by rotating the region about the x-axis has finite volume.

Hints

- Treat each integral as improper because the function is undefined at the left endpoint. - Find antiderivatives of \(x^{-1/2}\) and \(\frac{1}{x}\). - Compare the behavior of \(\sqrt{x}\) and \(\ln x\) as \(x\to0^+\).

Solution

1. The area is the improper integral \(A=\int_0^1x^{-1/2}\,\mathrm{d}x=\lim_{a\to0^+}[2\sqrt{x}]_a^1=2\). 2. The volume is \(V=\pi\int_0^1(g(x))^2\,\mathrm{d}x=\pi\int_0^1\frac{1}{x}\,\mathrm{d}x\). 3. \(V=\pi\lim_{a\to0^+}[\ln x]_a^1=\pi\lim_{a\to0^+}(0-\ln a)=\infty\). 4. Thus, the region has finite area, but its solid of revolution has infinite volume.

Answer

The area is \(2\) square units. The volume of the solid of revolution diverges and is infinite.

Disc method 53

A glass container has circular horizontal cross sections. Its bottom is at height \(x=-6\), and its rim is at \(x=0\). Its inside radius at height \(x\) is \(f(x)=\sqrt{\frac{12}{\pi}\cos\left(\frac{\pi x}{12}\right)}\). All lengths are in inches. a) Find the water volume when the water is \(3\,\text{in.}\) deep. Give an exact value and a value rounded to the nearest hundredth of a cubic inch. b) Adding \(4\,\text{in.}^3\) of water raises the water level from a depth \(h\) by exactly \(1\,\text{in.}\). Write an equation that can be used to find \(h\), where \(0\le h\le5\).

Hints

- Depth is measured upward from the bottom. How does that differ from the height coordinate \(x\)? - What area does a thin horizontal layer of water have? - In part b), what change in the volume function corresponds to one more inch of depth?

Solution

1. For water depth \(h\), \(V(h)=\pi\int_{-6}^{-6+h}[f(x)]^2\,\text{d}x=12\int_{-6}^{-6+h}\cos\left(\frac{\pi x}{12}\right)\,\text{d}x\). 2. Therefore \(V(h)=\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-6)}{12}\right)+1\right)\). 3. At \(h=3\), \(V(3)=\frac{144}{\pi}\left(1-\frac{\sqrt2}{2}\right)\,\text{in.}^3\approx13.43\,\text{in.}^3\). 4. The added water is the change \(V(h+1)-V(h)=4\). 5. Substituting gives \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\).

Answer

a) \(\frac{144}{\pi}\left(1-\frac{\sqrt2}{2}\right)\,\text{in.}^3\approx13.43\,\text{in.}^3\) b) \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\)

Disc method 54

A conical frustum has height \(h\). Its radius changes linearly from \(r\) at \(x=0\) to \(R\) at \(x=h\). Revolving the region between this radius graph and the x-axis gives the solid. Use integration to derive \(V=\frac13\pi h(R^2+Rr+r^2)\).

Hints

- First find the equation of the line through the two given endpoint radii. - What is the radius of a perpendicular disc at a general height? - Expand the squared radius before integrating. - Simplify the resulting polynomial in \(R\) and \(r\).

Solution

1. The line through \((0,r)\) and \((h,R)\) is \(f(x)=\frac{R-r}{h}x+r\). 2. By the disc method, \(V=\pi\int_0^h\left(\frac{R-r}{h}x+r\right)^2\,\text{d}x\). 3. Expanding and integrating gives \(V=\pi\left[\frac{(R-r)^2}{3h^2}x^3+\frac{r(R-r)}{h}x^2+r^2x\right]_0^h\). 4. Thus, \(V=\pi h\left(\frac{(R-r)^2}{3}+r(R-r)+r^2\right)\). 5. Simplifying yields \(V=\frac13\pi h(R^2+Rr+r^2)\).

Answer

\(V=\frac13\pi h(R^2+Rr+r^2)\)

Disc method 55

Let \(f(x)=\frac{1}{x+2}\) for \(x\ge0\). a) Show that the region between the graph of \(f\) and the x-axis over \([0,\infty)\) has infinite area. b) The region is rotated about the x-axis. Find the volume of the resulting unbounded solid.

Hints

- Identify the antiderivative needed for each integral. - How does the natural logarithm behave as its input increases without bound? - Compare the convergence of reciprocal first and second powers. - An unbounded region can still generate a solid with finite volume.

Solution

1. The area is \(A=\int_0^\infty\frac{1}{x+2}\,\mathrm{d}x=\lim_{b\to\infty}[\ln(x+2)]_0^b\). 2. Since \(\ln(b+2)-\ln2\to\infty\), the area diverges. 3. The volume is \(V=\pi\int_0^\infty\frac{1}{(x+2)^2}\,\mathrm{d}x\). 4. \(V=\pi\lim_{b\to\infty}[-\frac{1}{x+2}]_0^b=\pi\left(0+\frac{1}{2}\right)=\frac{\pi}{2}\).

Answer

a) The area is infinite. b) The volume is \(\frac{\pi}{2}\) cubic units.

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