0
$\begingroup$

So I want to find the volume when the region between $y=x^2$ and $y=4-x^2$ is rotated about the $x$-axis.

So I start by finding the roots where they meet, so I find :

$$\int_{ -\sqrt{2} }^{\sqrt{2}} \pi (4-x^2)^2 \, dx$$

But this got me the wrong answer $75.825$. Is that right or is the book wrong with $94.7$?

$\endgroup$
0
1
$\begingroup$

A typical cross-section is a washer, not a disk. So we must subtract the inner radius: $$ V = \pi\int_{ -\sqrt{2} }^{\sqrt{2}} [(4-x^2)^2 - (x^2)^2] \, dx = \frac{64\sqrt 2}{3}\pi = 94.7815\ldots $$

$\endgroup$
0
$\begingroup$

According to Mathematica you are both wrong:

volume integral

As Adriano says, you have to subtract the inner radius.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.