Can someone check my solution for AP Calc Volume of a Solid question

Question

Let $R$ be the region in the first quadrant bounded by $y=1/x$, the horizontal line, $y=1$, and the vertical line $x=e$.

Region $R$ is the base of a solid. For this solid, each cross section perpendicular to the y-axis is a semicircle. Find the volume of the solid.

Can someone please check my solution?

$A(y)=$ area of cross section $=$ area of semi circle with radius $=\frac{1}{2}\pi r^2$

When we graph these equations we get something like this:

The diameter of the semi circle is $e-\frac{1}{y}$

so the radius is $\frac{1}{2}(e-\frac{1}{y})$

Therefore the area is $A(y)=\frac{1}{2}\pi (\frac{1}{2}(e-\frac{1}{y}))^2 = \frac{\pi}{8}(e-\frac{1}{y})^2$

At $x=e$, $y=\frac{1}{e}$ and we are given that $y=1$ so we have $\frac{1}{e}\leq y \leq 1$

To find the volume, we have $V=\int_{a}^{b} A(y) dy$

Thus we have $V=\int_{1/e}^1 \frac{\pi}{8} (e-\frac{1}{y})^2 dy$

$V=\frac{\pi}{8}(e^2-2e-1)$

Answer 1

You have two mistakes. What you call diameter is actually the radius and what you call area is supposed to be a difference of squares while you wrote a square of a difference.

First the solid is like a half torus obtained by rotating the region R half turn around y-axis.

The volume is $\int_1^e{\pi\cdot x\cdot (1-\frac{1}{x})\cdot dx}$, where:

1. $\pi\cdot x$ is the semi-circular path described by the cross-section
2. $(1-\frac{1}{x})\cdot dx$ is the cross section

Numerically, the volume is $\pi\cdot\left[\frac{x^2}{2}-x\right ]_1^e=\frac{\pi}{2}\cdot(e-1)^2$