Free math worksheets

- 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.

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.

\(20\pi\) cubic units

- 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.

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.

\(3\) units

- 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 \(\pi\).

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. \(V=\pi[\frac{3}{2}x^2+x]_1^5=\pi(42.5-2.5)=40\pi\).

\(40\pi\) cubic units

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

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.

\(3\pi\) cubic units

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

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.

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

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

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.

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

- 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.

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 \(V=\pi\int_0^5[R(y)]^2\,\mathrm{d}y=35\pi\).

\(35\pi\) cubic units

- 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.

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.

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

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

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 \(9(12\pi)=108\pi\).

\(108\pi\) cubic units

- 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.

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 \(V=\pi\int_0^3(3-x)^2\,\mathrm{d}x\).

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

- For rotation about the x-axis, what vertical distance gives the disc radius? - The required response must include the disc-method integral before using any familiar-solid formula. - After evaluating the integral, identify the two end radii and the frustum height for the check.

1. The disc radius is \(R(x)=\frac{x}{2}\), so the required disc-method setup is \(V=\pi\int_k^{3k}\left(\frac{x}{2}\right)^2\,\mathrm{d}x\). 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 \(2k\), radii \(\frac{k}{2}\) and \(\frac{3k}{2}\). The frustum formula \(V=\frac{\pi h}{3}(R^2+Rr+r^2)\) also gives \(\frac{13\pi}{6}k^3\).

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

- 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.

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. \(V=\pi r[\frac{1}{2}x^2]_0^r=\frac{1}{2}\pi r^3\). 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 \(V_{\text{cyl}}=\pi r^3\). 4. Therefore, the solid of revolution has half the volume of the cylinder.

The solid has volume \(V=\frac{1}{2}\pi r^3\), which is half the cylinder's volume \(\pi r^3\).

- 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.

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 \(f(3)=5\).

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

- First write the paraboloid's volume as a function of \(a\). - Simplify \((\sqrt{x})^2\). - When squaring \(\frac{1}{2}x\), square both the coefficient and the variable. - Simplify the ratio by canceling common factors.

1. \(V_{\text{paraboloid}}=\pi\int_0^a(\sqrt{x})^2\,\text{d}x=\pi\int_0^a x\,\text{d}x=\frac{1}{2}\pi a^2\). Setting this equal to \(32\pi\) gives \(a^2=64\), so \(a=8\). 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. \(32\pi:\frac{128\pi}{3}=3:4\).

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

- 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.

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

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

- 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 \(h\)?

1. If the water reaches height \(h\), its volume is \(V(h)=\pi\int_0^h(e^{0.2x})^2\,\mathrm{d}x=\pi\int_0^h e^{0.4x}\,\mathrm{d}x\). 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 \(h=\frac{5}{2}\ln\left(1+\frac{24}{\pi}\right)\,\text{cm}\approx5.39\,\text{cm}\), which lies in the modeled interval.

\(\frac{5}{2}\ln\left(1+\frac{24}{\pi}\right)\,\text{cm}\approx5.39\,\text{cm}\)

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

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.

\(108\pi\) cubic units

- 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 \(x+1\).

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. \(V=\pi[-\frac{9}{x+1}]_0^2=\pi(-3+9)=6\pi\).

\(6\pi\) cubic units

- 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.

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. \(V=4\pi[-2e^{-0.5x}]_0^4=8\pi(1-e^{-2})\).

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

- Use the upper semicircle as the radius function for the discs. - Squaring \(\sqrt{R^2-x^2}\) removes the radical. - Use symmetry or evaluate the antiderivative at \(-R\) and \(R\).

1. Rotating the upper semicircle about the x-axis produces discs with radius \(\sqrt{R^2-x^2}\). 2. Therefore, \(V=\pi\int_{-R}^R(R^2-x^2)\,\mathrm{d}x\). 3. Evaluating gives \(V=\pi\left[R^2x-\frac{x^3}{3}\right]_{-R}^R=\frac{4}{3}\pi R^3\).

\(V=\frac{4}{3}\pi R^3\)

- 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 \(x\to0^+\).

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.

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

- 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.

1. By the disc method, \(V=\pi\int_{-4}^{4}(0.2x^2+4)\,\mathrm{d}x=\frac{608\pi}{15}\) model cubic units. 2. Since one model unit is \(10\,\text{cm}\), one model cubic unit is \(1000\,\text{cm}^3=1\,\text{L}\). Thus the stool 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, which is \(2\pi\,\text{L}\). 4. The remaining volume is \(\frac{608\pi}{15}-2\pi=\frac{578\pi}{15}\,\text{L}\approx121.06\,\text{L}\).

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

- 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 \(300\,\text{cm}^3\)?

1. At height \(8\), \(V(8)=\pi\int_0^8(0.2x+3)^2\,\mathrm{d}x\). 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 \(\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)=300\).

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

- Squaring the 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.

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

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

- 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 \(x\).

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, \(V=\pi\int_0^h\left(\frac{r}{h}x\right)^2\,\mathrm{d}x\). 3. Therefore, \(V=\pi\frac{r^2}{h^2}\left[\frac{x^3}{3}\right]_0^h=\frac{1}{3}\pi r^2h\).

a) \(f(x)=\frac{r}{h}x\) b) \(V=\frac{1}{3}\pi r^2h\)

- 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 \(x(y)\).

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, \(V=\frac{384\pi}{5}\,\text{cm}^3\approx241.27\,\text{cm}^3\).

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

- 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.

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 \(b=\frac{e^2-1}{2}\approx3.19\).

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

- 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.

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.

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

- Squaring the radius function removes the square root. - Use integration by parts for the product \(x\sin x\). - Recall the sine and cosine values at \(0\) and \(\pi\). - Include the factor \(\pi\) from the disc-method formula.

1. By the disc method, \(V=\pi\int_0^\pi(f(x))^2\,\mathrm{d}x=\pi\int_0^\pi x\sin x\,\mathrm{d}x\). 2. Use integration by parts with \(u=x\) and \(\mathrm{d}v=\sin x\,\mathrm{d}x\): \(\int x\sin x\,\mathrm{d}x=-x\cos x+\sin x\). 3. \(\int_0^\pi x\sin x\,\mathrm{d}x=[-x\cos x+\sin x]_0^\pi=\pi\). 4. Therefore, \(V=\pi(\pi)=\pi^2\) cubic units.

\(\pi^2\) cubic units

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

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

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

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

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

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

- 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.

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.

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

- 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.

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.

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

- 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.

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.

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

- 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.

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 \(V\approx257.8\,\text{cm}^3\).

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

- 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.

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, \(V\approx1389.2\,\text{cm}^3\).

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

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

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.

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

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

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.

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

- 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?

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.

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

- 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.

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.

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

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

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 \(k>0\), \(k=3\).

\(k=3\)

- 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.

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.

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

- 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.

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 \(V=\frac{9\pi}{16}\cdot\frac{64}{3}=12\pi\,\text{cm}^3\).

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

- Recall the disc-method formula for a solid formed by rotation about the x-axis. - Rewrite the improper integral using a variable upper limit. - When does \(x^n\) approach zero as \(x\to\infty\)? - Check separately the case in which the antiderivative is logarithmic.

1. Using the disc method, \(V=\pi\int_1^\infty(f_k(x))^2\,\mathrm{d}x=\pi\int_1^\infty x^{-2k}\,\mathrm{d}x\). 2. When \(k\ne\frac{1}{2}\), an antiderivative is \(\frac{x^{1-2k}}{1-2k}\). The improper integral converges only when \(1-2k<0\), or \(k>\frac{1}{2}\). 3. When \(k=\frac{1}{2}\), the integrand is \(x^{-1}\), and the integral diverges logarithmically. 4. For \(k>\frac{1}{2}\), \(\int_1^\infty x^{-2k}\,\mathrm{d}x=\frac{1}{2k-1}\). Therefore, \(V(k)=\frac{\pi}{2k-1}\).

The volume is finite for \(k>\frac{1}{2}\), and \(V(k)=\frac{\pi}{2k-1}\).

- 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 \(x\to0^+\).

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.

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

- A water depth of \(h\) places the water surface at \(x=-6+h\). - Simplify \(\pi[f(x)]^2\) before integrating. - Why does a fully submerged fixed object's displaced volume cancel when comparing two water levels?

1. For water depth \(h\), \(V(h)=\pi\int_{-6}^{-6+h}(f(x))^2\,\mathrm{d}x=12\int_{-6}^{-6+h}\cos\left(\frac{\pi x}{12}\right)\,\mathrm{d}x\). 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. Because the object is completely submerged both before and after the added water, its displaced volume cancels, so \(V(h+1)-V(h)=4\). 5. Substitution gives \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\).

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

- First find the equation of the line through the two given endpoint radii. - Use the disc-method formula. - Expand \((ax+b)^2\) before integrating. - Simplify the resulting polynomial in \(R\) and \(r\).

1. The line through \((0,r)\) and \((h,R)\) is \(f(x)=\frac{R-r}{h}x+r\). 2. By the disc method, \(V=\pi\int_0^h\left(\frac{R-r}{h}x+r\right)^2\,\mathrm{d}x\). 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 the expression in parentheses gives \(\frac{R^2+Rr+r^2}{3}\), so \(V=\frac{1}{3}\pi h(R^2+Rr+r^2)\).

\(V=\frac{1}{3}\pi h(R^2+Rr+r^2)\)

- 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.

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. \(V=\pi\lim_{b\to\infty}[-\frac{1}{x+2}]_0^b=\pi\left(0+\frac{1}{2}\right)=\frac{\pi}{2}\).

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