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

1. A solid is formed by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ around the line $y = 2$. What is the volume of this solid?
   Answer: $\frac{56\pi}{3}$ When revolving around $y = 2$, the outer radius is the distance from $y = 0$ to $y = 2$, which is 2. The inner radius is the distance from $y = \sqrt{x}$ to $y = 2$, which is $2 - \sqrt{x}$. Volume = $\pi \int_0^4 [2^2 - (2 - \sqrt{x})^2] dx = \pi \int_0^4 [4 - (4 - 4\sqrt{x} + x)] dx = \pi \int_0^4 [4\sqrt{x} - x] dx = \pi [\frac{8x^{3/2}}{3} - \frac{x^2}{2}]_0^4 = \pi [\frac{64}{3} - 8] = \pi [\frac{64 - 24}{3}] = \frac{56\pi}{3}$. Choice B forgets to expand $(2 - \sqrt{x})^2$ correctly. Choice C makes an integration error. Choice D uses the disk method incorrectly.
2. A region is bounded by $y = x^2 + 1$, $y = 5$, and the y-axis in the first quadrant. When this region is revolved around the x-axis, what is the volume of the solid formed?
   Answer: $\frac{256\pi}{15}$ First, find the intersection points: $x^2 + 1 = 5$ gives $x = 2$. The region extends from $x = 0$ to $x = 2$. Using the washer method, the outer radius is $R(x) = 5$ and inner radius is $r(x) = x^2 + 1$. Volume = $\pi \int_0^2 [5^2 - (x^2 + 1)^2] dx = \pi \int_0^2 [25 - x^4 - 2x^2 - 1] dx = \pi \int_0^2 [24 - x^4 - 2x^2] dx = \pi [24x - \frac{x^5}{5} - \frac{2x^3}{3}]_0^2 = \pi [48 - \frac{32}{5} - \frac{16}{3}] = \pi [\frac{720 - 192 - 80}{15}] = \frac{256\pi}{15}$. Choice A incorrectly uses disk method instead of washer. Choice C forgets to subtract the inner radius. Choice D makes an error in the polynomial expansion.

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