Free math worksheets

- Use the accumulated squared radii directly. - Subtract the inner contribution from the outer contribution before multiplying by the circular factor.

1. Washer volume is \(V=\pi\int_a^b([R(x)]^2-[r(x)]^2)\,\mathrm{d}x\). 2. By linearity, \(V=\pi(30-12)=18\pi\).

\(18\pi\) cubic units

- Read the two horizontal boundary heights from the graph. - Both radii are measured from the x-axis. - The farther boundary gives the outer radius.

1. The upper boundary is \(3\) units from the x-axis, so the outer radius is \(R=3\). 2. The lower boundary is \(1\) unit from the x-axis, so the inner radius is \(r=1\).

\(R=3\), \(r=1\)

- Which radius is farther from the x-axis throughout the interval? - Write washer area as outer circular area minus inner circular area. - Square each radius before subtracting.

1. A perpendicular slice produces a washer with outer radius \(R(x)=\sqrt{x+2}\) and inner radius \(r(x)=0.2x+1\). 2. Therefore, \(V=\pi\int_0^{10}[(x+2)-(0.2x+1)^2]\,\mathrm{d}x\). 3. Evaluating gives \(V=70\pi-\frac{130\pi}{3}=\frac{80\pi}{3}\,\text{cm}^3\).

\(\frac{80\pi}{3}\,\text{cm}^3\)

- Which boundary is farther from the x-axis? - Use outer radius squared minus inner radius squared. - Find the x-value where the two boundaries meet.

1. The curves meet at \(x=16\). On \([0,16]\), the outer radius is \(4\) and the inner radius is \(\sqrt{x}\). 2. Thus, \(V=\pi\int_0^{16}[4^2-(\sqrt{x})^2]\,\mathrm{d}x=\pi\int_0^{16}(16-x)\,\mathrm{d}x\). 3. Therefore, \(V=128\pi\) cubic units.

\(128\pi\) cubic units

- Which function is farther from the x-axis on the interval? - Subtract the squared radii. - Rewrite the reciprocal square with a negative exponent before integrating.

1. On \([1,2]\), the outer radius is \(\frac{1}{x}\) and the inner radius is \(0.5\). 2. Therefore, \(V=\pi\int_1^2\left(\frac{1}{x^2}-0.25\right)\,\mathrm{d}x=\frac{\pi}{4}\) cubic units.

\(\frac{\pi}{4}\) cubic units

- Compare the two boundaries by their distances from the x-axis. - Washer area must be nonnegative on every slice. - Which squared radius must be subtracted from the other?

1. On \([0,3]\), \(4-x\ge1\), so the outer radius is \(4-x\) and the inner radius is \(1\). 2. The correct integral is \(V=\pi\int_0^3[(4-x)^2-1]\,\mathrm{d}x\). 3. Evaluating gives \(V=18\pi\) cubic units.

\(18\pi\) cubic units

- Compare distances to each axis rather than comparing the absolute y-coordinates. - Shift the region and the axis together. - If every cross-sectional area is unchanged, the accumulated volume is unchanged.

1. In the original solid, the outer and inner radii are \(f(x)\) and \(g(x)\). 2. In the shifted solid, the distances to \(y=5\) are again \(f(x)\) and \(g(x)\). 3. The washer areas are identical at every \(x\), with the same limits. 4. Therefore, the new volume is \(V\).

The new volume is \(V\).

- Measure each radius from the dashed axis rather than from the x-axis. - Which displayed line is farther from the axis of rotation? - Washer area is outer radius squared minus inner radius squared, multiplied by \(\pi\).

1. The upper boundary is \(y=x+1\), so its distance from \(y=-1\) is the outer radius \(R(x)=x+2\). 2. The lower boundary is \(y=x\), so its distance from \(y=-1\) is the inner radius \(r(x)=x+1\). 3. Therefore, \(A(x)=\pi[(x+2)^2-(x+1)^2]\).

\(A(x)=\pi[(x+2)^2-(x+1)^2]\)

- In each panel, check whether the shaded slice touches the axis of rotation. - A slice touching the axis has no central hole. - Measure all radii as perpendicular distances from the panel's axis of rotation.

1. In panel a), the lower boundary \(y=1\) is the axis of rotation, so each slice produces a disc with radius \((3-x)-1=2-x\). Thus, \(A_a(x)=\pi(2-x)^2\). 2. In panel b), the region lies above the axis \(y=0\), so each slice produces a washer with outer radius \(3-x\) and inner radius \(1\). Thus, \(A_b(x)=\pi[(3-x)^2-1]\).

a) Disc: \(A_a(x)=\pi(2-x)^2\) b) Washer: \(A_b(x)=\pi[(3-x)^2-1]\)

- Measure both radii from the x-axis. - For a washer, which circular areas must be subtracted? - Evaluate the thickness-squared expression separately so you can compare the two models.

1. The cross sections are washers with outer radius \(R(x)=\sqrt{x}+1\) and inner radius \(r(x)=1\). 2. Thus, \(V=\pi\int_0^4[(\sqrt{x}+1)^2-1^2]\,\mathrm{d}x=\frac{56\pi}{3}\) cubic units. 3. The comparison integral is \(V_{\text{incorrect}}=\pi\int_0^4(\sqrt{x})^2\,\mathrm{d}x=8\pi\) cubic units. 4. The values differ because washer area is \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).

a) \(\frac{56\pi}{3}\) cubic units b) \(8\pi\) cubic units; it is smaller because \(R^2-r^2\ne(R-r)^2\).

- Determine which function gives the outer radius on the interval. - Use the washer-method formula. - Simplify the integrand before integrating. - Expand the squared binomial carefully.

1. On \([1,3]\), \(f(x)\ge g(x)\ge0\), so the washer method gives \(V=\pi\int_1^3(f(x)^2-g(x)^2)\,\mathrm{d}x\). 2. \(V=\pi\int_1^3((x+1)^2-(\sqrt{2x+2})^2)\,\mathrm{d}x=\pi\int_1^3(x^2-1)\,\mathrm{d}x\). 3. \(V=\pi[\frac{x^3}{3}-x]_1^3=\pi\left(6+\frac{2}{3}\right)=\frac{20\pi}{3}\).

\(V=\frac{20\pi}{3}\) cubic units

- Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers with negative exponents before integrating. - Multiply the integral by \(\pi\).

1. On \([1,2]\), \(f(x)\ge g(x)\ge0\), so \(f\) is the outer radius and \(g\) is the inner radius. 2. \(V=\pi\int_1^2\left(\frac{1}{x^2}-\frac{1}{x^4}\right)\,\mathrm{d}x=\pi\int_1^2(x^{-2}-x^{-4})\,\mathrm{d}x\). 3. \(V=\pi[-x^{-1}+\frac{1}{3}x^{-3}]_1^2=\pi\left(-\frac{11}{24}+\frac{2}{3}\right)=\frac{5\pi}{24}\).

\(V=\frac{5\pi}{24}\) cubic units

- Treat the tower as an exterior solid with an interior cavity removed. - Why do the exterior and interior volume integrals begin at different x-values? - Keep the outer and inner squared-radius contributions separate until the subtraction is clear.

1. The full exterior volume is \(V_{\text{out}}=\pi\int_0^{60}4.5x\,\mathrm{d}x=8100\pi\,\text{m}^3\). 2. The interior cavity begins at \(x=0.8\), so \(V_{\text{in}}=\pi\int_{0.8}^{60}4.2(x-0.8)\,\mathrm{d}x=7359.744\pi\,\text{m}^3\). 3. The material volume is \(V_{\text{out}}-V_{\text{in}}=740.256\pi\,\text{m}^3\approx2325.58\,\text{m}^3\).

\(740.256\pi\,\text{m}^3\approx2325.58\,\text{m}^3\)

- First find the container volume up to a variable water depth. - A solid object reduces the space available for water. - Determine whether the cylinder is partially or fully submerged at each depth. - Subtract the displaced cylinder volume from the container volume.

1. The container volume up to height \(h\) is \(V(h)=\pi\int_0^h(f(x))^2\,\mathrm{d}x=\pi\int_0^h(2x+4)\,\mathrm{d}x=\pi(h^2+4h)\). 2. At \(h=2\), the container volume is \(12\pi\,\text{in.}^3\). The cylinder occupies \(\pi(1)^2(2)=2\pi\,\text{in.}^3\), so the water volume is \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\). 3. At \(h=5\), the container volume is \(45\pi\,\text{in.}^3\). The entire cylinder remains submerged and displaces \(2\pi\,\text{in.}^3\), so the water volume is \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\).

a) \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\) b) \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\)

- Find the intersection points first. - Determine which function gives the outer radius. - Use the washer-method formula. - Expand the squared expressions carefully.

1. The curves intersect where \(x^2=4x-x^2\), giving \(x=0\) and \(x=2\). 2. On \([0,2]\), \(g(x)\ge f(x)\ge0\), so the washer method gives \(V=\pi\int_0^2(g(x)^2-f(x)^2)\,\mathrm{d}x\). 3. \((4x-x^2)^2-(x^2)^2=16x^2-8x^3\). 4. \(V=\pi[\frac{16}{3}x^3-2x^4]_0^2=\frac{32\pi}{3}\).

\(V=\frac{32\pi}{3}\) cubic units

- Find the intersection points to determine the bounds. - Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers using negative exponents. - Check that the resulting volume is positive.

1. The curves intersect where \(\frac{2}{x}=3-x\), which gives \(x^2-3x+2=0\). Thus, the bounds are \(x=1\) and \(x=2\). 2. On \([1,2]\), \(g(x)\ge f(x)>0\), so \(V=\pi\int_1^2\left((3-x)^2-\left(\frac{2}{x}\right)^2\right)\,\mathrm{d}x\). 3. \(V=\pi\int_1^2(x^2-6x+9-4x^{-2})\,\mathrm{d}x\). 4. \(V=\pi[\frac{x^3}{3}-3x^2+9x+\frac{4}{x}]_1^2=\pi\left(\frac{32}{3}-\frac{31}{3}\right)=\frac{\pi}{3}\).

\(V=\frac{\pi}{3}\) cubic units

- Identify the outer and inner radii of each washer. - Subtract squared radii before integrating. - Keep the exact value until the final decimal approximation.

1. Each perpendicular cross section is a washer with outer radius \(R(x)=4-0.1x^2\) and inner radius \(r(x)=2\). 2. Thus \(V=\pi\int_0^3[(4-0.1x^2)^2-2^2]\,\mathrm{d}x\). 3. Evaluating gives \(V=\frac{14643\pi}{500}\,\text{m}^3\approx92.00\,\text{m}^3\).

\(\frac{14643\pi}{500}\,\text{m}^3\approx92.00\,\text{m}^3\)

- View the bowl as an outer solid with an inner cavity removed. - Why do the outer and inner profiles have different starting x-values? - The bowl's capacity is the volume generated by the inner radius.

1. The outer volume is \(V_{\text{outer}}=\pi\int_0^{16}(0.5x+4)\,\mathrm{d}x=128\pi\,\text{cm}^3\). 2. The capacity is the inner volume: \(V_{\text{inner}}=\pi\int_4^{16}(0.8x-3.2)\,\mathrm{d}x=57.6\pi\,\text{cm}^3\approx180.96\,\text{cm}^3\). 3. Therefore, the glass volume is \(V_{\text{outer}}-V_{\text{inner}}=70.4\pi\,\text{cm}^3\approx221.17\,\text{cm}^3\).

The glass volume is \(70.4\pi\,\text{cm}^3\approx221.17\,\text{cm}^3\). The capacity is \(57.6\pi\,\text{cm}^3\approx180.96\,\text{cm}^3\).

- Subtract the inner cross-sectional area from the outer cross-sectional area. - Apply exponent rules when squaring each radius function. - Use the antiderivative of \(e^{kx}\). - The two integrals can be combined because they use the same interval.

1. Use washers: \(V=\pi\int_0^{10}(f(x)^2-g(x)^2)\,\mathrm{d}x\). 2. \(f(x)^2-g(x)^2=4e^{0.2x}-2.25e^{0.2x}=1.75e^{0.2x}\). 3. Therefore, \(V=1.75\pi\int_0^{10}e^{0.2x}\,\mathrm{d}x=1.75\pi[5e^{0.2x}]_0^{10}\). 4. Thus, \(V=8.75\pi(e^2-1)\approx175.63\,\text{cm}^3\).

\(8.75\pi(e^2-1)\,\text{cm}^3\approx175.63\,\text{cm}^3\)

- Find the intersection points to determine the washer bounds. - Which function is farther from the x-axis inside the interval? - Apply a power-reduction identity to the squared sine term.

1. The curves intersect at \(x=0\) and \(x=\frac{\pi}{2}\). 2. On this interval, \(\sin x\) is the outer radius. Thus, \(V=\pi\int_0^{\pi/2}\left[\sin^2x-\left(\frac{2x}{\pi}\right)^2\right]\,\mathrm{d}x\). 3. The component integrals are \(\int_0^{\pi/2}\sin^2x\,\mathrm{d}x=\frac{\pi}{4}\) and \(\int_0^{\pi/2}\frac{4x^2}{\pi^2}\,\mathrm{d}x=\frac{\pi}{6}\). 4. Therefore, \(V=\frac{\pi^2}{12}\) cubic units.

\(\frac{\pi^2}{12}\) cubic units

- Solve for the intersection points and check any solutions introduced by squaring. - Identify which function gives the outer radius. - Use the washer-method formula. - Subtract the squares of the functions, not the square of their difference.

1. Set the functions equal: \(\sqrt{x+4}=\frac{1}{2}x+2\). Squaring and checking the solutions gives intersections at \(x=-4\) and \(x=0\). 2. On \([-4,0]\), \(f(x)\ge g(x)\), and both functions are nonnegative. Thus, \(V=\pi\int_{-4}^{0}\left((\sqrt{x+4})^2-\left(\frac{x}{2}+2\right)^2\right)\,\mathrm{d}x\). 3. The integrand simplifies to \(-\frac{x^2}{4}-x\). 4. \(V=\pi\left[-\frac{x^3}{12}-\frac{x^2}{2}\right]_{-4}^{0}=\pi\left(\frac{8}{3}\right)=\frac{8\pi}{3}\approx8.38\) cubic units.

\(\frac{8\pi}{3}\) cubic units, or approximately \(8.38\) cubic units

- Find the intersection points to determine the interval. - Identify the outer and inner radii. - Square each function separately before subtracting. - A sketch can help confirm which graph is farther from the x-axis.

1. Solve \(x^2-4x+5=-x+5\). This gives \(x(x-3)=0\), so the curves intersect at \(x=0\) and \(x=3\). 2. On \([0,3]\), \(g(x)\ge f(x)>0\), so \(g\) is the outer radius and \(f\) is the inner radius. 3. \(V=\pi\int_0^3(g(x)^2-f(x)^2)\,\mathrm{d}x\). 4. The difference is \((-x+5)^2-(x^2-4x+5)^2=-x^4+8x^3-25x^2+30x\). 5. \(V=\pi\left[-\frac{x^5}{5}+2x^4-\frac{25x^3}{3}+15x^2\right]_0^3=\frac{117\pi}{5}\approx73.51\) cubic units.

\(\frac{117\pi}{5}\) cubic units, or approximately \(73.51\) cubic units

- Find the intersection points first. - For area, integrate the upper function minus the lower function. - For volume, identify the outer and inner radii. - Square each radius function before subtracting.

1. Solving \(\frac{1}{2}x^2=x+4\) gives intersection points \(x=-2\) and \(x=4\). 2. On \([-2,4]\), \(g(x)\ge f(x)\). The area is \(A=\int_{-2}^{4}\left(x+4-\frac{x^2}{2}\right)\,\mathrm{d}x=18\) square units. 3. Both functions are nonnegative on the interval, so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 4. \(V=\pi\int_{-2}^{4}\left((x+4)^2-\left(\frac{x^2}{2}\right)^2\right)\,\mathrm{d}x\). 5. \(V=\pi\left[\frac{x^3}{3}+4x^2+16x-\frac{x^5}{20}\right]_{-2}^{4}=\frac{576\pi}{5}\approx361.91\) cubic units.

a) \(18\) square units b) \(\frac{576\pi}{5}\) cubic units, or approximately \(361.91\) cubic units

- Set the functions equal to find the limits. - Use the logarithmic antiderivative for \(1/x\). - For the volume, subtract the squares of the radii. - Rewrite \(1/x^2\) as a negative power before integrating.

1. Solving \(4-x=\frac{3}{x}\) gives \(x^2-4x+3=0\), so the curves intersect at \(x=1\) and \(x=3\). 2. On \([1,3]\), \(h(x)\ge k(x)\). Thus, \(A=\int_1^3\left(4-x-\frac{3}{x}\right)\,\mathrm{d}x\). 3. \(A=\left[4x-\frac{x^2}{2}-3\ln x\right]_1^3=4-3\ln3\approx0.704\) square units. 4. Using washers, \(V=\pi\int_1^3\left((4-x)^2-\left(\frac{3}{x}\right)^2\right)\,\mathrm{d}x\). 5. \(V=\pi\left[16x-4x^2+\frac{x^3}{3}+\frac{9}{x}\right]_1^3=\frac{8\pi}{3}\approx8.378\) cubic units.

a) \(4-3\ln3\) square units, or approximately \(0.704\) square units b) \(\frac{8\pi}{3}\) cubic units, or approximately \(8.378\) cubic units

- Find the intersection points first. - Identify the outer and inner radii. - Expand both squared binomials carefully. - Symmetry can shorten the calculation.

1. Solving \(x^2+1=9-x^2\) gives \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(g(x)\ge f(x)>0\), so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 3. \(V=\pi\int_{-2}^{2}\left((9-x^2)^2-(x^2+1)^2\right)\,\mathrm{d}x\). 4. The integrand simplifies to \(80-20x^2\). 5. Thus, \(V=\pi\left[80x-\frac{20x^3}{3}\right]_{-2}^{2}=\frac{640\pi}{3}\approx670.21\) cubic units.

\(\frac{640\pi}{3}\) cubic units, or approximately \(670.21\) cubic units

- Set the function values equal and check any solutions introduced by squaring. - Which graph is farther from the x-axis on the enclosed interval? - Subtract squared radii before integrating.

1. Set the functions equal: \(\sqrt{2x}=x\). Squaring and checking gives \(x=0\) and \(x=2\), so the intersection points are \((0,0)\) and \((2,2)\). 2. On \([0,2]\), \(f(x)\ge g(x)\). Using washers, \(V=\pi\int_0^2(2x-x^2)\,\mathrm{d}x\). 3. Therefore, \(V=\pi\left[x^2-\frac{x^3}{3}\right]_0^2=\frac{4\pi}{3}\) cubic units.

a) \((0,0)\) and \((2,2)\) b) \(\frac{4\pi}{3}\) cubic units

- Use the inner profile to find the cavity volume. - The delayed start of the inner profile creates a solid glass base. - Subtract the cavity volume from the full outer volume, then use mass equals density times volume.

1. The cavity volume is \(V_{\text{cavity}}=\pi\int_1^6(4x-4)\,\mathrm{d}x=50\pi\,\text{cm}^3\approx157.1\,\text{cm}^3=157.1\,\text{mL}\). 2. The full outer volume is \(V_{\text{outer}}=\pi\int_0^6(0.5x+3)^2\,\mathrm{d}x=126\pi\,\text{cm}^3\). 3. Therefore, the glass volume is \(76\pi\,\text{cm}^3\approx238.8\,\text{cm}^3\), and the mass is \(2.5(76\pi)=190\pi\,\text{g}\approx596.9\,\text{g}\).

a) \(157.1\,\text{mL}\) b) \(76\pi\,\text{cm}^3\approx238.8\,\text{cm}^3\) c) \(596.9\,\text{g}\)

- Convert the hole's diameter to its radius. - What are the outer and inner radii of a perpendicular washer? - Subtract the squared radii before integrating over the spool's length.

1. The hole has radius \(1\,\text{cm}\), so each perpendicular cross section is a washer with outer radius \(f(x)\) and inner radius \(1\). 2. Thus, \(V=\pi\int_0^8\left([f(x)]^2-1^2\right)\,\mathrm{d}x=\pi\int_0^8\left(0.5(x-4)^2+3\right)\,\mathrm{d}x\). 3. Evaluating gives \(V=\frac{136\pi}{3}\,\text{cm}^3\).

\(\frac{136\pi}{3}\,\text{cm}^3\)

- Solve for the first-quadrant intersections. - Which curve gives the outer radius on the enclosed interval? - Square both radius functions before subtracting.

1. The curves intersect at \(x=0\) and \(x=1\). 2. On \([0,1]\), \(x^2\ge x^4\). Thus, \(V=\pi\int_0^1(x^4-x^8)\,\mathrm{d}x\). 3. Therefore, \(V=\frac{4\pi}{45}\) cubic units.

\(\frac{4\pi}{45}\) cubic units

- Use the intersections as the integration limits. - Which function gives the outer radius on the enclosed interval? - Square both radii before subtracting.

1. The curves intersect at \(x=0\) and \(x=2\). 2. On \([0,2]\), \(2x\ge x^2\). Thus, \(V=\pi\int_0^2(4x^2-x^4)\,\mathrm{d}x\). 3. Therefore, \(V=\frac{64\pi}{15}\) cubic units.

\(\frac{64\pi}{15}\) cubic units

- Use the intersections as the integration limits. - Which curve is farther from the x-axis on the enclosed interval? - Expand the squared binomial after forming outer radius squared minus inner radius squared.

1. The curves intersect at \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(f(x)\) is the outer radius and \(1\) is the inner radius. Thus, \(V=\pi\int_{-2}^{2}[(-x^2+5)^2-1]\,\mathrm{d}x\). 3. Expanding and using symmetry gives \(V=2\pi\int_0^2(x^4-10x^2+24)\,\mathrm{d}x=\frac{832\pi}{15}\) cubic units.

\(\frac{832\pi}{15}\) cubic units

- Use the inner profile for capacity and the outer profile for total volume. - How does the solid base from \(x=0\) to \(x=1\) affect the glass volume? - After finding the glass volume, use the given density for the mass.

1. The capacity is \(V_{\text{inside}}=\pi\int_1^4(2\sqrt{x-1})^2\,\mathrm{d}x=18\pi\,\text{cm}^3\). 2. The total outer volume is \(V_{\text{outer}}=\pi\int_0^4(0.2x+3)^2\,\mathrm{d}x=\frac{3484\pi}{75}\,\text{cm}^3\). 3. The glass volume is \(\frac{2134\pi}{75}\,\text{cm}^3\), so the mass is \(2.5\cdot\frac{2134\pi}{75}=\frac{1067\pi}{15}\,\text{g}\). 4. At the top, \(f(4)-g(4)=3.8-2\sqrt3\,\text{cm}\), the wall thickness at the rim.

a) \(18\pi\,\text{cm}^3\) b) \(\frac{2134\pi}{75}\,\text{cm}^3\) c) \(\frac{1067\pi}{15}\,\text{g}\) d) The wall thickness at the top rim is \((3.8-2\sqrt3)\,\text{cm}\).

- Use horizontal distances from the y-axis as the washer radii. - Which boundary is farther from the y-axis throughout the interval? - Subtract the squared inner radius from the squared outer radius.

1. A horizontal slice has outer radius \(R(y)=5-y\) and inner radius \(r(y)=\frac{y^2}{4}\). 2. Therefore, \(V=\pi\int_0^2\left[(5-y)^2-\left(\frac{y^2}{4}\right)^2\right]\,\mathrm{d}y\). 3. Evaluating gives \(V=\frac{484\pi}{15}\) cubic units.

\(\frac{484\pi}{15}\) cubic units

- Measure both radii from the shifted axis, not from the x-axis. - Which line is farther from \(y=-2\)? - Subtract the squared distances before integrating.

1. The outer radius is \(R(x)=(x+1)-(-2)=x+3\), and the inner radius is \(r(x)=x-(-2)=x+2\). 2. Therefore, \(V=\pi\int_0^3[(x+3)^2-(x+2)^2]\,\mathrm{d}x=24\pi\) cubic units.

\(24\pi\) cubic units

- Translate the constant boundaries into constant radii. - Use the interval restriction to select the valid root.

1. The washers have outer radius \(5\) and inner radius \(k\). 2. \(V=\pi\int_0^4(25-k^2)\,\mathrm{d}x=4\pi(25-k^2)\). 3. Set \(4\pi(25-k^2)=64\pi\), so \(k^2=9\). 4. The restriction gives \(k=3\).

\(k=3\)

- Keep the unknown in the upper integration limit. - Simplify the difference of squared radii before integrating.

1. \(V=\pi\int_0^b[(2x)^2-x^2] \,\mathrm{d}x=\pi b^3\). 2. Set \(\pi b^3=21\pi\). 3. Thus \(b=\sqrt[3]{21}\).

\(b=\sqrt[3]{21}\)

- Compare the two pairs of distances to the axes. - A reflected placement can preserve both washer radii.

1. About \(y=0\), the washer area is \(\pi(5^2-2^2)=21\pi\). 2. About \(y=7\), the distances are \(5\) and \(2\) again, so the washer area is also \(21\pi\). 3. Both solids have the same length \(6\), so their volumes are equal.

The volumes are equal.

- Confirm which graph is farther from the axis on the whole interval. - Keep the difference of squared radii inside the numerical integral.

1. The upper curve gives the outer radius on \([0,1]\). 2. \(V=\pi\int_0^1\left[(2+e^{-x^2})^2-\left(1+\frac{x}{2}\right)^2\right] \,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx18.856\).

\(\approx18.856\) cubic units

- For an axis above the region, the lower boundary can be farther away. - Use symmetry only after forming the washer area correctly.

1. The outer radius is the distance from \(y=3\) to \(y=0\), so \(R=3\). 2. The inner radius is \(r(x)=3-x^2\). 3. \(V=\pi\int_{-1}^1[9-(3-x^2)^2] \,\mathrm{d}x=\frac{18\pi}{5}\).

\(\frac{18\pi}{5}\) cubic units

- Rewrite both boundaries as x-values in terms of \(y\). - Use the intersection heights as the integration limits. - Which horizontal distance from the y-axis is larger for \(0<y<4\)?

1. The curves meet at heights \(y=0\) and \(y=4\). Rewrite them as \(x=\sqrt{y}\) and \(x=\frac{y}{2}\). 2. A horizontal slice has outer radius \(\sqrt{y}\) and inner radius \(\frac{y}{2}\). 3. Therefore, \(V=\pi\int_0^4\left(y-\frac{y^2}{4}\right)\,\mathrm{d}y=\frac{8\pi}{3}\) cubic units.

\(\frac{8\pi}{3}\) cubic units

- The two radii share the same variable factor. - Use the stated range to select the physically valid parameter.

1. The outer radius is \(4x\), and the inner radius is \(kx\). 2. \(V=\pi\int_0^1(16-k^2)x^2\,\mathrm{d}x=\frac{\pi}{3}(16-k^2)\). 3. Set \(\frac{\pi}{3}(16-k^2)=4\pi\), giving \(k^2=4\). 4. The restriction gives \(k=2\).

\(k=2\)

- Write the area of a washer as the area of the outer circle minus the inner circle. - Compare that expression with the expansion of \((a-b)^2\). - Under what conditions are \(a^2-b^2\) and \((a-b)^2\) equal? - What does an inner radius of zero mean geometrically?

1. The actual washer area is \(A(x)=\pi(f(x)^2-g(x)^2)\). 2. The proposed expression is \(\pi(f(x)-g(x))^2=\pi(f(x)^2-2f(x)g(x)+g(x)^2)\). 3. Setting the expressions equal and simplifying gives \(2g(x)(f(x)-g(x))=0\). 4. Therefore, equality holds only when \(g(x)=0\), so the cross-section is a disc, or when \(f(x)=g(x)\), so the cross-sectional area is zero. 5. For a genuine washer with \(f(x)>g(x)>0\), the correct integrand is \(\pi(f(x)^2-g(x)^2)\). Squaring the radial thickness \(f(x)-g(x)\) ignores the washer's distance from the axis.

\(\pi(f^2-g^2)-\pi(f-g)^2=2\pi g(f-g)\). This difference is positive when \(f>g>0\), so \(\pi(f-g)^2\) cannot be used for a genuine washer. The correct volume uses \(\pi\int(f^2-g^2)\,\mathrm{d}x\).

- Use horizontal slices because the axis of rotation is vertical. - For an axis to the right of the region, which boundary is farther from the axis? - Write each radius as a positive horizontal distance from \(x=6\).

1. The axis is to the right of both curves. The left boundary is farther from the axis, so \(R(y)=6-y^2\). 2. The inner radius is \(r(y)=6-(2y+1)=5-2y\). 3. Therefore, \(V=\pi\int_0^1[(6-y^2)^2-(5-2y)^2] \,\mathrm{d}y=\frac{238\pi}{15}\) cubic units.

\(\frac{238\pi}{15}\) cubic units

- Check whether each vertical slice stays on one side of the x-axis. - What kind of cross section results when a slice crosses the axis of rotation? - Where does the lower boundary change from negative to nonnegative?

1. On \([0,1]\), the vertical slice crosses the x-axis, so rotating it produces a disc of radius \(2\), not a washer with an inner radius. 2. On \([1,2]\), the slice remains above the x-axis and produces a washer with outer radius \(2\) and inner radius \(x-1\). 3. Thus, \(V=\pi\int_0^1 4\,\mathrm{d}x+\pi\int_1^2[4-(x-1)^2]\,\mathrm{d}x=\frac{23\pi}{3}\). 4. The proposed value is \(V_s=\frac{22\pi}{3}\), so it underestimates the volume by \(\frac{\pi}{3}\).

The proposal is incorrect. The correct volume is \(\frac{23\pi}{3}\) cubic units, and \(V_s\) is too small by \(\frac{\pi}{3}\) cubic units.

- Check whether the vertical axis lies inside each horizontal slice. - At what y-value does the left boundary reach \(x=1\)? - Measure every radius as a horizontal distance from the axis \(x=1\).

1. For \(0\le y\le1\), the horizontal slice crosses \(x=1\), producing a disc of radius \(3-y\). 2. For \(1\le y\le2\), both boundaries lie to the right of the axis, producing a washer with outer radius \(3-y\) and inner radius \(y-1\). 3. Thus, \(V=\pi\int_0^1(3-y)^2\,\mathrm{d}y+\pi\int_1^2[(3-y)^2-(y-1)^2] \,\mathrm{d}y\). 4. The two contributions are \(\frac{19\pi}{3}\) and \(2\pi\), so \(V=\frac{25\pi}{3}\) cubic units.

\(\frac{25\pi}{3}\) cubic units