Compute the volumes of solids bounded by the following surfaces:
$$x^2+y^2-2ax=0$$ $$ z=0$$ $$x^2+y^2=z^2$$
how can i solve this with multiple integrals i try changing variable but I got to $\int_{0}^{\pi}\int_{0}^{\sqrt{a}}\sqrt{(r\cos\theta +a)^2+r^2\sin^2\theta}$ i can't solve it .
the answer from book is ${32\over 9 }a^3$
Thanks in advance!
Not sure what your change of coordinates might have been. But here is what it looks like for me.
Convert to polar:
$x = r \cos \theta\\ y = r \sin \theta\\ z = z\\ dx\;dy\;dz = r\;dz\;dr\;d\theta$
Substitute into the original equations to find our limits.
$r^2 \cos^2 \theta + r^2 \sin^2\theta - 2ar \cos\theta =0\\ r = 2a\cos\theta$
Since this cylindar is to the right of the $y$ axis, $\theta$ from -$\frac \pi2$ to $\frac \pi 2$ is appropriate
$z = 0$
$r^2 \cos^2 \theta + r^2 \sin^2\theta = z^2\\ z = r$
$\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\int_0^{2a\cos\theta}\int_0^r r\;dz\;dr\;d\theta$
Suppose you went
$x = r \cos \theta+a\\ y = r \sin \theta\\ z = z\\ dx\;dy\;dz = r\;dz\;dr\;d\theta$
$r = a$
$r^2 + 2ra\cos\theta + a^2= z^2$
$\int_{0}^{2\pi}\int_0^{a}\int_0^{\sqrt{r^2+2ra\cos\theta+a^2}} r\;dz\;dr\;d\theta$
That looks close to what you have.
$\int_{0}^{2\pi}\int_0^{a} r\sqrt{r^2+2ra\cos\theta+a^2}\;dr\;d\theta$
you dropped the Jacobian...but, this is indeed a messier intergral.