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Find the volume of the solid of revolution obtained by rotating the region bounded by

$f(x) = x^3 + 1$, $g(x) = x^2$ and $0 ≤ x ≤ 1$ about the line $y = 3$.

I know the gist of the problem, but I'm struggling with the $y = 3$. If it was simply rotating around the $x$ axis I would just plug in the outer and inner radius into $\pi r^2$, but when $y$ is set equal to a number I'm not sure how to handle the problem. An explanation would be helpful.

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    $\begingroup$ Hint: the $x$ axis is the axis $y=0$. Rotating around $y=3$ is similar... $\endgroup$
    – Martigan
    Oct 23, 2014 at 14:27

2 Answers 2

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If you look at the graph of $f(x), g(x)$, and eyeball the line $y = 3$,

enter image description here

you'll see the inner radius will be $r_i = 3-(x^3 + 1) = 2-x^3$ and the outer radius will be $r_o = 3-x^2$.

Then our integrand will be of the form $\pi\big(r_0^2 - r_i^2)$.

That gives us the integral $$\pi \int_0^1 \left((3-x^2)^2 - (2 - x^3)^2\right)\,dx$$

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  • $\begingroup$ So if y = -3, would it be 3+ (x^3 + 1) and 3 + x^2? $\endgroup$
    – user3032755
    Oct 23, 2014 at 14:38
  • $\begingroup$ Yes, but the outer radius will be $3+y^3 + 1$, inner radius $x^2$, with respect to $y = -3$. $\endgroup$
    – amWhy
    Oct 23, 2014 at 14:40
  • $\begingroup$ oops, I meant outer radius would be $3+x^3 +1$, inner radius $x^2$ with respect to rotating the target area about $y= -3$ $\endgroup$
    – amWhy
    Oct 23, 2014 at 18:56
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When you intersect the described rotational body $B$ with a plane $x={\rm const.}$ you obtain an annulus in this plane of outer radius $3-x^2$ and inner radius $3-(x^3+1)=2-x^3$. The area $A(x)$ of this annulus is given by $$A(x)=\pi\bigl((3-x^2)^2-(2-x^3)^2\bigr)\qquad(0\leq x\leq 1)\ .$$ Therefore the volume of $B$ is given by $$V(B)=\int_0^1 A(x)\ dx={142\pi\over35}\ .$$

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