How to account for solids of revolution around vertical lines to the right of the x axis?

Question

I'm trying to find the volume of a solid created by rotating the region enclosed between $x=y^2$ and $x=1$ around the line $x=8$.

Noting that the intersections of the functions occur at $(0,0)$ and $(1,1)$, I can see that $y=\sqrt{x}$ is the upper function.

I tried using the formula $$\pi\int_0^1{((1+7)^2-(x^2+7)^2)dx}$$

However this doesn't yield the correct answer. What am I doing wrong?

Answer 1

Using the disc method, you have that $V=\int_{-1}^1 π(R^2-r^2)dy$ where $R=8-y^2$ and $r=8-1$,

so $\displaystyle V=\int_{-1}^1 \pi((8-y^2)^2-7^2)\;dy$.

Alternatively, using the shell method,

$\;\;\displaystyle V=\int_0^12\pi r(x)h(x)dx=\int_0^12\pi (8-x)(2\sqrt{x})\;dx$

Answer 2

You are rotating about a vertical line. So you can use

• the "discs" or "washers" method, integrating with respect to $y$; or
• the "cylindrical shells" method, integrating with respect to $x$.

But you are trying to use the "discs" method and integrating with respect to $x$.

Hint: draw a picture!

Answer 3

The area being rotated is $(1;8)\times(0;\sqrt x)$ or $(0;8)\times(0;1)\cup(1;y^2)\times(1;\sqrt 8)$

This is being rotated around the vertical line $x=8$, so either use $2\pi \int_1^8 r~y\operatorname d x$ or $\pi(7^2+\int_1^{\sqrt 8} r^2\operatorname d y)$.

Can you figure out what $r$ is in terms of $x$ and $y$ respectively?