Washer Method: x- or y-Axis - Calculus 1

1. Let $R$ be the region in the first quadrant bounded by the graphs of $y=x^2+1$, $y=5$, and $x=1$. What is the volume of the solid generated by revolving $R$ about the y-axis?
   Answer: $\frac{9\pi}{2}$ To revolve around the y-axis, we must integrate with respect to $y$. The region is bounded by $x=1$ on the left, $y=5$ on the top, and $y=x^2+1$ on the bottom-right. We need to express $x$ in terms of $y$: $y=x^2+1 \implies x=\sqrt{y-1}$ (since we are in the first quadrant). The bounds for $y$ are from the intersection of $x=1$ and $y=x^2+1$ (which is $y=1^2+1=2$) up to $y=5$. So, we integrate from $y=2$ to $y=5$. For a given $y$ in this interval, the outer radius is the distance from the y-axis to the curve $x=\sqrt{y-1}$, so $R(y)=\sqrt{y-1}$. The inner radius is the distance from the y-axis to the line $x=1$, so $r(y)=1$. The volume is: $V = \pi \int_{2}^{5} [(\sqrt{y-1})^2 - 1^2] dy = \pi \int_{2}^{5} (y-1-1) dy = \pi \int_{2}^{5} (y-2) dy$ $V = \pi \left[ \frac{y^2}{2} - 2y \right]_2^5 = \pi \left[ (\frac{25}{2}-10) - (\frac{4}{2}-4) \right] = \pi \left[ \frac{5}{2} - (-2) \right] = \frac{9\pi}{2}$ Distractor A results from using incorrect bounds, such as integrating from 1 to 5. Distractor B is the volume of the solid using the disk method with radius $R(y)=\sqrt{y-1}$, ignoring the hole created by revolving around $x=1$. Distractor D is the volume of the solid of revolution for the region bounded by $y=x^2+1$ and $y=5$ (without the $x=1$ boundary).
2. Let R be the region enclosed by the graphs of $f(x) = \sqrt{x}$ and $g(x) = x^2$. What is the volume of the solid generated by revolving R about the x-axis?
   Answer: $\frac{3\pi}{10}$ To find the volume using the washer method, we integrate $\pi(R(x)^2 - r(x)^2)$ with respect to x. First, find the points of intersection by setting $\sqrt{x} = x^2$, which gives $x=0$ and $x=1$. These are the limits of integration. For $x \in [0, 1]$, $\sqrt{x} \ge x^2$, so the outer radius is $R(x) = \sqrt{x}$ and the inner radius is $r(x) = x^2$. The volume integral is $V = \pi \int_0^1 ((\sqrt{x})^2 - (x^2)^2) dx = \pi \int_0^1 (x - x^4) dx$. Evaluating the integral gives $\pi [\frac{1}{2}x^2 - \frac{1}{5}x^5]_0^1 = \pi (\frac{1}{2} - \frac{1}{5}) = \frac{3\pi}{10}$.
3. Let $k$ be a positive constant. The region bounded by the parabola $y=kx^2$ and the line $y=k$ is revolved about the x-axis. If the volume of the resulting solid is $\frac{16\pi}{5}$, what is the value of $k$?
   Answer: $\sqrt{2}$ First, find the bounds of integration by setting the functions equal: $kx^2=k \implies x^2=1 \implies x=\pm 1$. The line $y=k$ is the upper boundary, so the outer radius is $R(x)=k$. The parabola $y=kx^2$ is the lower boundary, so the inner radius is $r(x)=kx^2$. The volume $V$ is given by: $V = \pi \int_{-1}^{1} [k^2 - (kx^2)^2] dx = \pi \int_{-1}^{1} (k^2 - k^2x^4) dx$ $V = \pi k^2 \int_{-1}^{1} (1 - x^4) dx = \pi k^2 \left[ x - \frac{x^5}{5} \right]_{-1}^{1}$ $V = \pi k^2 \left( (1 - \frac{1}{5}) - (-1 - \frac{-1}{5}) \right) = \pi k^2 \left( \frac{4}{5} - (-\frac{4}{5}) \right) = \frac{8\pi k^2}{5}$ We are given that the volume is $\frac{16\pi}{5}$. So, we set them equal: $\frac{8\pi k^2}{5} = \frac{16\pi}{5} \implies 8k^2 = 16 \implies k^2 = 2$ Since $k$ is positive, $k=\sqrt{2}$. Distractor C, $k=2$, might arise from an algebraic error such as $4k=8$. Distractor B arises from other miscalculations. Distractor A results from a different algebraic mistake.
4. The region bounded by $y=x^2+2$ and $y=6$ is revolved about the x-axis. What is the volume of the resulting solid?
   Answer: $\frac{1408\pi}{15}$ When you encounter a solid of revolution problem, you're using the disk or washer method to find volume. Since this region is bounded by two curves and revolved about the x-axis, you'll need the washer method because there's a hollow center. First, find where the curves intersect by setting $x^2 + 2 = 6$, which gives $x^2 = 4$, so $x = \pm 2$. The region extends from $x = -2$ to $x = 2$, with $y = 6$ as the outer curve and $y = x^2 + 2$ as the inner curve. For the washer method, the volume is $V = \pi \int_{-2}^{2} [R(x)^2 - r(x)^2] \, dx$, where $R(x) = 6$ (outer radius) and $r(x) = x^2 + 2$ (inner radius). $V = \pi \int_{-2}^{2} [6^2 - (x^2 + 2)^2] \, dx = \pi \int_{-2}^{2} [36 - (x^4 + 4x^2 + 4)] \, dx$ $V = \pi \int_{-2}^{2} (32 - 4x^2 - x^4) \, dx$ Evaluating: $V = \pi \left[32x - \frac{4x^3}{3} - \frac{x^5}{5}\right]_{-2}^{2}$ Since the integrand is even, you can compute from 0 to 2 and double it: $V = 2\pi \left[64 - \frac{32}{3} - \frac{32}{5}\right] = 2\pi \left[\frac{320 - 160 - 96}{15}\right] = \frac{1408\pi}{15}$ Answer C is correct. Answer A likely comes from using only the outer radius, B from calculation errors in the integration, and D from forgetting to expand $(x^2 + 2)^2$ properly. Remember: always identify which curve is outer/inner, and be careful expanding binomial expressions in washer problems.
5. A solid is generated by revolving the region bounded by $y=x^2$, $y=4$, and $x=0$ about the x-axis. What is the volume of the solid?
   Answer: $\frac{128\pi}{5}$ The region is bounded by $x=0$ and the intersection of $y=x^2$ and $y=4$, which is $x=2$. For revolution about the x-axis, the outer radius is the distance from the axis (y=0) to the outer curve (y=4), so $R(x)=4$. The inner radius is the distance from the axis to the inner curve ($y=x^2$), so $r(x)=x^2$. The volume is $V = \pi \int_0^2 (4^2 - (x^2)^2) dx = \pi \int_0^2 (16 - x^4) dx$. Evaluating gives $\pi [16x - \frac{1}{5}x^5]_0^2 = \pi (32 - \frac{32}{5}) = \frac{128\pi}{5}$.
6. The region bounded by the parabola $x=(y-2)^2$ and the line $x=4$ is revolved about the y-axis. Which of the following integrals correctly represents the volume of the generated solid?
   Answer: $\pi \int_{0}^{4} (16 - (y-2)^4) dy$ The revolution is about the y-axis, so we integrate with respect to $y$. Find the intersection points: $4 = (y-2)^2 \implies \pm 2 = y-2 \implies y=0$ and $y=4$. These are the bounds of integration. For any $y$ in $[0,4]$, the line $x=4$ is farther from the y-axis than the parabola $x=(y-2)^2$. Thus, the outer radius is $R(y)=4$ and the inner radius is $r(y)=(y-2)^2$. Using the washer method formula $V = \pi \int_{c}^{d} [R(y)^2 - r(y)^2] dy$, we get: $V = \pi \int_{0}^{4} [4^2 - ((y-2)^2)^2] dy = \pi \int_{0}^{4} [16 - (y-2)^4] dy$ Distractor A uses the incorrect formula $\pi \int (R-r)^2 dy$. Distractor B is the integral for the volume using the shell method for revolution about the x-axis. Distractor D incorrectly swaps the outer and inner radii.
7. The region enclosed by $y=x^2+1$ and $y=x+3$ is rotated about the x-axis. The volume of the solid generated is given by which expression?
   Answer: $\pi \int_{-1}^2 (-x^4 - x^2 + 6x + 8) dx$ First, find the intersection points by solving $x^2+1 = x+3$, which simplifies to $x^2-x-2=0$, or $(x-2)(x+1)=0$. The limits of integration are $x=-1$ and $x=2$. On this interval, the line $y=x+3$ is above the parabola $y=x^2+1$. Thus, the outer radius is $R(x)=x+3$ and the inner radius is $r(x)=x^2+1$. The integrand for the washer method is $R(x)^2 - r(x)^2 = (x+3)^2 - (x^2+1)^2 = (x^2+6x+9) - (x^4+2x^2+1) = -x^4 - x^2 + 6x + 8$. The complete integral is $\pi \int_{-1}^2 (-x^4 - x^2 + 6x + 8) dx$.
8. The region in the first quadrant bounded by $y=1/x$, $x=1$, and $y=4$ is revolved about the y-axis. Which integral represents the volume of the solid?
   Answer: $\pi \int_1^4 (1 - 1/y^2) dy$ When you encounter a volume of revolution problem, the key is identifying which axis of rotation to use and setting up the correct method. Since we're revolving around the y-axis, you'll want to express everything in terms of y and use the washer method. First, sketch the region. The curve $y = 1/x$ intersects the line $x = 1$ at point $(1,1)$, and the horizontal line $y = 4$ intersects the curve at $x = 1/4$. So our region extends from $y = 1$ to $y = 4$. For revolution about the y-axis, each horizontal slice creates a washer with outer radius and inner radius. At height $y$, the outer radius extends from the y-axis to the line $x = 1$, giving radius = 1. The inner radius extends from the y-axis to the curve $x = 1/y$ (solving $y = 1/x$ for $x$), giving radius = $1/y$. Using the washer formula $V = \pi \int [R^2 - r^2] dy$, we get $\pi \int_1^4 (1^2 - (1/y)^2) dy = \pi \int_1^4 (1 - 1/y^2) dy$. Choice A has the radii reversed, giving a negative volume. Choice C attempts to use x as the variable of integration, which doesn't match y-axis rotation. Choice D uses $(1 - 1/y)$ instead of the correct $(1 - 1/y^2)$, forgetting to square the radii. Remember: for y-axis rotation, integrate with respect to y, and always square both the outer and inner radii in the washer method.
9. Let $R$ be the region enclosed by the graphs of $y = x+2$ and $y = x^2$. What is the volume of the solid generated by revolving $R$ about the x-axis? Give your answer in simplest form.
   Answer: $\frac{72\pi}{5}$ First, find the points of intersection by setting the two equations equal: $x^2 = x+2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$. The points of intersection are $x=-1$ and $x=2$. In the interval $[-1, 2]$, the line $y=x+2$ is above the parabola $y=x^2$. When revolving around the x-axis, the outer radius is $R(x) = x+2$ and the inner radius is $r(x) = x^2$. The volume $V$ is given by the washer method formula: $V = \pi \int_{-1}^{2} [R(x)^2 - r(x)^2] dx = \pi \int_{-1}^{2} [(x+2)^2 - (x^2)^2] dx$ $V = \pi \int_{-1}^{2} (x^2 + 4x + 4 - x^4) dx$ $V = \pi \left[ -\frac{x^5}{5} + \frac{x^3}{3} + 2x^2 + 4x \right]_{-1}^{2}$ $V = \pi \left( \left(-\frac{32}{5} + \frac{8}{3} + 8 + 8\right) - \left(\frac{1}{5} - \frac{1}{3} + 2 - 4\right) \right)$ $V = \pi \left( \left(-\frac{32}{5} + \frac{8}{3} + 16\right) - \left(\frac{1}{5} - \frac{1}{3} - 2\right) \right)$ $V = \pi \left( \frac{-96+40+240}{15} - \frac{3-5-30}{15} \right) = \pi \left( \frac{184}{15} - \frac{-32}{15} \right) = \frac{216\pi}{15} = \frac{72\pi}{5}$ Distractor A results from an arithmetic error in the final step. Distractor C may result from calculation errors, for example, incorrectly evaluating the definite integral. Distractor D is the unsimplified fraction.
10. The region R is bounded by the graphs of $y=2\sqrt{x}$ and $y=x$. Which integral gives the volume of the solid generated by revolving R about the x-axis?
    Answer: $\pi \int_0^4 (4x - x^2) dx$ When finding the volume of a solid of revolution about the x-axis, you use the disk method: $V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$, where R(x) is the outer radius and r(x) is the inner radius. First, find where the curves intersect by solving $2\sqrt{x} = x$. Squaring both sides: $4x = x^2$, so $x^2 - 4x = 0$, giving $x = 0$ and $x = 4$. Between these points, $2\sqrt{x} > x$ (you can verify by testing $x = 1$), so $y = 2\sqrt{x}$ is the upper curve. When revolving about the x-axis, the outer radius is the distance from the x-axis to the upper curve: $R(x) = 2\sqrt{x}$. The inner radius is the distance to the lower curve: $r(x) = x$. The volume integral becomes: $V = \pi \int_0^4 [(2\sqrt{x})^2 - x^2] \, dx = \pi \int_0^4 (4x - x^2) \, dx$ This matches answer choice D. Choice A uses dy integration, which would require the washer method with horizontal slices—but the setup is incorrect. Choice B uses $(2\sqrt{x} - x)^2$, which represents the area between curves squared, not the difference of areas of circles. Choice C has the terms reversed: it subtracts the larger radius squared from the smaller one, giving a negative volume. Strategy tip: Always identify which curve is on top, set up your radii correctly, and remember the disk/washer method uses the difference of the squares of the radii, not the square of their difference.
11. Let R be the region bounded by $x=y^2$ and $x=4$. What is the volume of the solid generated when R is revolved about the y-axis?
    Answer: $\frac{256\pi}{5}$ When you encounter a volume of revolution problem, you need to identify the region, the axis of rotation, and choose the appropriate method. Since we're revolving around the y-axis and the region is naturally described in terms of y, the washer method works perfectly here. First, visualize the region R. The parabola $x = y^2$ opens rightward, and the vertical line $x = 4$ bounds it. These curves intersect where $y^2 = 4$, so $y = \pm 2$. The region extends from $y = -2$ to $y = 2$. For revolution about the y-axis, each horizontal slice creates a washer with outer radius 4 (distance from y-axis to the line $x = 4$) and inner radius $y^2$ (distance from y-axis to the parabola). The volume formula becomes: $V = \pi \int_{-2}^{2} (4^2 - (y^2)^2) \, dy = \pi \int_{-2}^{2} (16 - y^4) \, dy$ $V = \pi \left[16y - \frac{y^5}{5}\right]_{-2}^{2} = \pi \left[\left(32 - \frac{32}{5}\right) - \left(-32 + \frac{32}{5}\right)\right]$ $V = \pi \left[64 - \frac{64}{5}\right] = \pi \cdot \frac{256}{5} = \frac{256\pi}{5}$ Answer A ($64\pi$) likely comes from using just the outer radius without accounting for the hollow center. Answer B ($\frac{128\pi}{5}$) suggests integrating from 0 to 2 instead of -2 to 2. Answer D ($\frac{512\pi}{5}$) might result from incorrectly squaring terms in the setup. Remember: always sketch the region first, identify which variable to integrate with respect to, and carefully set up your radii based on the axis of rotation.
12. Let R be the region bounded by $y=x$, the x-axis, the y-axis, and the line $x=k$ for some $k>0$. Let S be the region bounded by $y=x$, the y-axis, and the line $y=k$. If S is revolved around the y-axis and R is revolved around the x-axis, the resulting solids have the same volume. This statement is true for:
    Answer: all $k>0$ When you encounter volumes of revolution problems, you're applying the disk/washer method or cylindrical shell method to find volumes of 3D solids formed by rotating 2D regions around axes. Let's find the volumes systematically. For region R (bounded by $y=x$, x-axis, y-axis, and $x=k$), when revolved around the x-axis, we use the disk method: $V_R = \pi \int_0^k x^2 \, dx = \pi \left[\frac{x^3}{3}\right]_0^k = \frac{\pi k^3}{3}$ For region S (bounded by $y=x$, y-axis, and $y=k$), when revolved around the y-axis, we again use the disk method. Since $y=x$ means $x=y$, the radius at height $y$ is $y$: $V_S = \pi \int_0^k y^2 \, dy = \pi \left[\frac{y^3}{3}\right]_0^k = \frac{\pi k^3}{3}$ Both volumes equal $\frac{\pi k^3}{3}$, so they're equal for all $k>0$. Why the wrong answers fail: Choice A) suggests no value works, but we just showed they're always equal. Choice B) claims only $k=1$ works, missing that the relationship holds universally. Choice C) proposes $k=\sqrt[3]{3}$ as the unique solution, but this stems from incorrectly setting up or solving the volume equations. The key insight here is recognizing the symmetry between these two setups. Both regions are essentially the same triangle oriented differently, and the revolution methods preserve this geometric relationship. Always set up your integrals carefully and look for underlying symmetries that might make problems simpler than they initially appear.
13. Let R be the region bounded by $y=\ln(x)$, $x=e$, and the x-axis. Which of the following integrals gives the volume of the solid generated when R is revolved about the y-axis?
    Answer: $\pi \int_0^1 (e^2 - e^{2y}) dy$ For revolution about the y-axis, we integrate with respect to y. The region is bounded below by $y=0$. The upper y-bound is where $x=e$ intersects $y=\ln(x)$, which is $y=\ln(e)=1$. The limits are $y=0$ to $y=1$. The outer radius is the distance from the y-axis to the line $x=e$, so $R(y)=e$. The inner radius is the distance from the y-axis to the curve $y=\ln(x)$, which must be expressed as $x=e^y$, so $r(y)=e^y$. The volume is $V = \pi \int_0^1 (R(y)^2 - r(y)^2) dy = \pi \int_0^1 (e^2 - (e^y)^2) dy = \pi \int_0^1 (e^2 - e^{2y}) dy$.
14. Let R be the region in the first quadrant bounded by $x=y^2$ and $x=2y$. Which integral represents the volume of the solid formed by revolving R about the y-axis?
    Answer: $\pi \int_0^2 (4y^2 - y^4) dy$ For revolution about the y-axis, we integrate with respect to y. First, find the intersection points: $y^2 = 2y \implies y(y-2)=0$, so $y=0$ and $y=2$. These are the limits. On the interval $y \in [0,2]$, $2y \ge y^2$. Thus, the outer radius is $R(y) = 2y$ and the inner radius is $r(y) = y^2$. The volume integral is $V = \pi \int_0^2 (R(y)^2 - r(y)^2) dy = \pi \int_0^2 ((2y)^2 - (y^2)^2) dy = \pi \int_0^2 (4y^2 - y^4) dy$.
15. Let R be the region bounded by the parabola $x = (y-2)^2$ and the line $x=4$. Which integral represents the volume of the solid formed by revolving R about the y-axis?
    Answer: $\pi \int_0^4 (16 - (y-2)^4) dy$ To revolve about the y-axis, we integrate with respect to y. Find intersection points by setting $(y-2)^2 = 4$, which gives $y-2 = \pm 2$, so $y=0$ and $y=4$. These are the integration limits. The axis of revolution is the y-axis ($x=0$). The outer boundary is the line $x=4$, so $R(y)=4$. The inner boundary is the parabola $x=(y-2)^2$, so $r(y)=(y-2)^2$. The volume integral is $V = \pi \int_0^4 (R(y)^2 - r(y)^2) dy = \pi \int_0^4 (4^2 - ((y-2)^2)^2) dy = \pi \int_0^4 (16 - (y-2)^4) dy$.
16. What is the volume of the solid generated by revolving the region bounded by $y=e^x$, $y=1$, and $x=1$ about the x-axis?
    Answer: $\frac{\pi}{2}(e^2-3)$ The region is bounded by $y=e^x$ (top), $y=1$ (bottom), and $x=1$ (right). The intersection of $y=e^x$ and $y=1$ is at $e^x=1$, which means $x=0$. So, the integration is from $x=0$ to $x=1$. The outer radius is $R(x)=e^x$ and the inner radius is $r(x)=1$. The volume is: $V = \pi \int_{0}^{1} [(e^x)^2 - 1^2] dx = \pi \int_{0}^{1} (e^{2x} - 1) dx$ $V = \pi \left[ \frac{1}{2}e^{2x} - x \right]_{0}^{1}$ $V = \pi \left( (\frac{1}{2}e^2 - 1) - (\frac{1}{2}e^0 - 0) \right) = \pi \left( \frac{1}{2}e^2 - 1 - \frac{1}{2} \right) = \pi(\frac{e^2}{2} - \frac{3}{2}) = \frac{\pi}{2}(e^2-3)$ Distractor B is the result of using the disk method with radius $e^x$ and ignoring the hole, i.e., forgetting to subtract the volume of the inner cylinder. Distractor C results from an error in evaluating the integral at the lower bound, forgetting the $-(\frac{1}{2}-0)$ term. Distractor D is the result of using the incorrect formula $\pi \int (R-r)^2 dx$.
17. The region enclosed by the parabolas $y = 2-x^2$ and $y=x^2$ is revolved about the x-axis. What is the volume of the solid generated?
    Answer: $\frac{16\pi}{3}$ First, find the points of intersection: $x^2 = 2-x^2 \implies 2x^2=2 \implies x^2=1 \implies x = \pm 1$. The region is symmetric about the y-axis. For $x$ in $[-1, 1]$, the parabola $y=2-x^2$ is above $y=x^2$. Thus, the outer radius is $R(x)=2-x^2$ and the inner radius is $r(x)=x^2$. The volume is: $V = \pi \int_{-1}^{1} [(2-x^2)^2 - (x^2)^2] dx$ $V = \pi \int_{-1}^{1} [(4 - 4x^2 + x^4) - x^4] dx = \pi \int_{-1}^{1} (4 - 4x^2) dx$ Using symmetry, we can write: $V = 2\pi \int_{0}^{1} (4 - 4x^2) dx = 2\pi \left[ 4x - \frac{4}{3}x^3 \right]_{0}^{1}$ $V = 2\pi \left( (4 - \frac{4}{3}) - 0 \right) = 2\pi \left( \frac{8}{3} \right) = \frac{16\pi}{3}$ Distractor A might result from forgetting to use symmetry and only integrating from 0 to 1, or another calculation error. Distractor C or D would result from incorrect algebraic expansion or integration.
18. The region bounded by $y=e^x$, $y=1$, and $x=2$ is revolved about the x-axis. What is the volume of the resulting solid?
    Answer: $\pi(\frac{e^4-1}{2} - 2)$ The region is bounded by $x=0$ (where $y=e^x$ intersects $y=1$) and $x=2$. For revolution about the x-axis, the outer radius is $R(x) = e^x$ and the inner radius is $r(x)=1$. The volume is $V = \pi \int_0^2 ((e^x)^2 - 1^2) dx = \pi \int_0^2 (e^{2x} - 1) dx$. Evaluating the integral gives $\pi [\frac{1}{2}e^{2x} - x]_0^2 = \pi ((\frac{1}{2}e^4 - 2) - (\frac{1}{2}e^0 - 0)) = \pi (\frac{e^4}{2} - 2 - \frac{1}{2}) = \pi(\frac{e^4-1}{2} - 2)$.
19. The region in the first quadrant bounded by $y=2\sqrt{x}$, the line $x=a$ (for some $a>0$), and the x-axis is revolved about the y-axis. The volume of this solid is found to be equal to the volume of the solid generated by revolving the same region about the x-axis. What is the value of $a$?
    Answer: $\frac{25}{16}$ First, find the volume $V_x$ from revolving about the x-axis. This is a disk method problem. The radius is $R(x)=2\sqrt{x}$. $V_x = \pi \int_0^a (2\sqrt{x})^2 dx = \pi \int_0^a 4x dx = \pi [2x^2]_0^a = 2\pi a^2$ Next, find the volume $V_y$ from revolving about the y-axis. We must use the washer method. The curve is $x = y^2/4$. The region is bounded on the right by $x=a$ and on the left by $x=y^2/4$. The y-values range from $0$ to $2\sqrt{a}$. The outer radius is $R(y)=a$ and the inner radius is $r(y)=y^2/4$. $V_y = \pi \int_0^{2\sqrt{a}} [a^2 - (y^2/4)^2] dy = \pi \int_0^{2\sqrt{a}} (a^2 - \frac{y^4}{16}) dy$ $V_y = \pi \left[ a^2y - \frac{y^5}{80} \right]_0^{2\sqrt{a}} = \pi \left( a^2(2\sqrt{a}) - \frac{(2\sqrt{a})^5}{80} \right) = \pi \left( 2a^{5/2} - \frac{32a^{5/2}}{80} \right) = \pi \left( 2a^{5/2} - \frac{2}{5}a^{5/2} \right) = \frac{8\pi}{5}a^{5/2}$ Now, set $V_x = V_y$: $2\pi a^2 = \frac{8\pi}{5}a^{5/2}$ Since $a>0$, we can divide by $2\pi a^2$: $1 = \frac{4}{5}a^{1/2} \implies \sqrt{a} = \frac{5}{4} \implies a = \frac{25}{16}$ Distractor A is the value of $\sqrt{a}$, not $a$. Distractor C is the reciprocal of the correct answer. Distractor B is from an algebraic error in solving for $\sqrt{a}$.
20. The region bounded by $x=y^2$ and $x=y+2$ is revolved about the y-axis. Which integral represents the volume of the solid?
    Answer: $\pi \int_{-1}^2 ((y+2)^2 - y^4) dy$ When finding volumes of solids of revolution about the y-axis, you need to determine whether to use the disk/washer method or cylindrical shells. Since this region is bounded by two curves and we're revolving about the y-axis, the washer method is most natural here. First, find where the curves intersect by solving $y^2 = y + 2$, which gives $y^2 - y - 2 = 0$, so $(y-2)(y+1) = 0$. The curves intersect at $y = -1$ and $y = 2$. Between these y-values, the line $x = y + 2$ is always to the right of the parabola $x = y^2$ (the outer radius), while $x = y^2$ forms the inner radius. Using the washer method, the volume is $V = \pi \int_{-1}^2 [R(y)^2 - r(y)^2] dy$, where $R(y) = y + 2$ (outer radius) and $r(y) = y^2$ (inner radius). This gives $V = \pi \int_{-1}^2 [(y+2)^2 - (y^2)^2] dy = \pi \int_{-1}^2 [(y+2)^2 - y^4] dy$, which is answer D. Option A squares the entire difference $(y+2-y^2)^2$, which would be incorrect for the washer method. Option B has the radii reversed, making the inner radius larger than the outer radius. Option C uses the shell method formula $2\pi \int y \cdot h(y) dy$, but this approach is more complex here since you'd need to split the integral due to the curves' orientations. Strategy tip: For revolution about the y-axis with vertically simple regions, always check which radius is larger at each y-value, then apply the washer formula: outer radius squared minus inner radius squared.

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