Disc Method: x- or y-Axis - Calculus 2

1. Consider the region in the first quadrant bounded by $y = x^2$, $y = 2x$, and the y-axis. When this region is revolved around the y-axis, what integral represents the volume using the disc method?
   Answer: $\pi \int_0^4 \left[\left(\frac{y}{2}\right)^2 - (\sqrt{y})^2\right] \, dy$ The curves intersect when $x^2 = 2x$, giving $x = 0$ and $x = 2$, so $y$ ranges from 0 to 4. For revolution around the y-axis, we need the washer method (a form of disc method) since there's a hole. The outer radius is from $y = 2x$ (so $x = \frac{y}{2}$) and the inner radius is from $y = x^2$ (so $x = \sqrt{y}$). The volume integral is $\pi \int_0^4 \left[\left(\frac{y}{2}\right)^2 - (\sqrt{y})^2\right] \, dy$. Choice A forgets the inner radius. Choice C reverses the outer and inner radii. Choice D uses the wrong variable of integration and incorrect setup.
2. The curve $x = y^2 - 4$ intersects the y-axis at two points. The region between this curve and the y-axis is revolved around the y-axis. What is the volume of the resulting solid?
   Answer: $\frac{128\pi}{5}$ The curve $x = y^2 - 4$ intersects the y-axis when $x = 0$, so $0 = y^2 - 4$, giving $y = \pm 2$. For $-2 \leq y \leq 2$, the curve is to the left of the y-axis. When revolved around the y-axis, we get discs with radius $r = |y^2 - 4| = 4 - y^2$ (since $y^2 \leq 4$ in this interval). The volume is $V = \pi \int_{-2}^{2} (4 - y^2)^2 \, dy = \pi \int_{-2}^{2} (16 - 8y^2 + y^4) \, dy = \pi \left[16y - \frac{8y^3}{3} + \frac{y^5}{5}\right]_{-2}^{2} = \pi \left[32 - \frac{64}{3} + \frac{32}{5}\right] \cdot 2 = \frac{128\pi}{5}$. Other choices result from sign errors or incorrect radius calculations.
3. The area between $y = \sqrt{4-x^2}$ and the x-axis for $-2 \leq x \leq 2$ is revolved around the x-axis. If the volume is $k\pi$, what is the value of $k$?
   Answer: $\frac{32}{3}$ When you encounter a problem involving revolving a region around an axis, you're dealing with the disk/washer method for finding volumes of revolution. The key is identifying the radius function and setting up the correct integral. First, recognize that $y = \sqrt{4-x^2}$ is the upper half of a circle with radius 2 centered at the origin. When you revolve this semicircle around the x-axis, you create a solid sphere with radius 2. Using the disk method, the volume is $V = \pi \int_{-2}^{2} [R(x)]^2 \, dx$, where $R(x) = \sqrt{4-x^2}$ is the radius at each x-value. This gives us: $V = \pi \int_{-2}^{2} (\sqrt{4-x^2})^2 \, dx = \pi \int_{-2}^{2} (4-x^2) \, dx$ Evaluating: $V = \pi \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \pi \left[\left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right)\right] = \pi \left[16 - \frac{16}{3}\right] = \frac{32\pi}{3}$ Therefore, $k = \frac{32}{3}$, which is answer D. Answer A ($\frac{16}{3}$) likely comes from forgetting to account for the negative portion of the interval. Answer B ($16$) might result from incorrectly computing $4x$ over the interval without the $x^3$ term. Answer C ($8$) could come from various computational errors in the integration or evaluation. Remember: when revolving around the x-axis, always square the y-function inside your integral, and be careful with your interval arithmetic when evaluating definite integrals.
4. The region bounded by $y = e^x$, $y = 1$, and $x = \ln 3$ is revolved around the line $y = 1$. Using the disc method, what integral represents the volume?
   Answer: $\pi \int_0^{\ln 3} (e^x - 1)^2 \, dx$ When revolving around $y = 1$, the radius of each disc is the distance from the curve $y = e^x$ to the line $y = 1$, which is $r = e^x - 1$. The region extends from $x = 0$ (where $e^0 = 1$) to $x = \ln 3$. Therefore, the volume integral is $\pi \int_0^{\ln 3} (e^x - 1)^2 \, dx$. Choice B attempts to use $y$ as the variable but has an incorrect radius formula. Choice C forgets to account for the shift to $y = 1$. Choice D uses the wrong variable and wrong radius.
5. The region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ is revolved around the x-axis. If the resulting solid is sliced by planes perpendicular to the x-axis, and the cross-sectional area at position $x = a$ is $9\pi$ square units, what is the value of $a$?
   Answer: $a = 9$ When the region is revolved around the x-axis, cross-sections perpendicular to the x-axis are circular discs with radius $r = \sqrt{x}$. The area of such a disc is $A(x) = \pi r^2 = \pi(\sqrt{x})^2 = \pi x$. Setting this equal to $9\pi$: $\pi x = 9\pi$, so $x = 9$. Therefore $a = 9$. Choice A results from incorrectly taking the square root of 9. Choice C results from forgetting to divide out the $\pi$ factor. Choice D results from incorrectly solving $\sqrt{x} = 9\pi$ instead of $\pi x = 9\pi$.

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