Finding volume of solid using shell method.

Question

Rotating the region bounded by $ y = x^3 , y = 0 , x=2$ around the line $ y = 8 $

I just want to double check if my initial formula is on the right track.

$$V_y = \int_0^8 (8-y)(y^{\frac{1}{3}})dy $$

Am I on the right track?

Answer 1

Since you are rotating around the line $y=8$, which is parallel to the $x$-axis, it makes sense to integrate along the $x$-axis (i.e., integrate $dx$) rather than along $y$. The outer radius at the point $x$ (let's call it $R(x)$) will be from $y=8$ to $y=0$ and the inner radius at the point $x$ (let's call it $r(x)$) will be from $y=8$ to $y=x^3$; this all happens over the region where $x$ goes from $0$ to $2$. So, $$ \int_0^2 \pi [R(x)^2 - r(x)^2] \,dx = \int_0^2 \pi [(8-0)^2 - (8-x^3)^2] \,dx = \cdots $$

Answer 2

Given values are $y=x^3,y=0,x=2$ rotating on $ y=8$. rotating axis $y = 8$ is parallel to the x-axis, so we have to apply the formula for rotating with $y = k$; \begin{equation} V_y(x) = 2\pi \int_a^b (y-k) f(y) dy \end{equation} Thus, the radius is $8-y$ with thickness $dy$, and height $f(y)$ (you can easily verify using graph with reference https://en.wikipedia.org/wiki/Shell_integration. Indeed, the height is unchanged.) Therefore, \begin{equation} V_y(x) = 2\pi \int_0^8 (8-y)(y^{\frac{1}{3}} ) dy \end{equation} The minus sign comes from that the rotated graph is under the rotating axis $y = 8$.

Answer 3

You are on the right track, but I think you want

$V=\int_0^8 2\pi r(y)h(y)dy =\int_0^8 2\pi (8-y)(2-y^{1/3})dy$.