Integrating over a paraboloid, help on determining limits of integrations

Question

Calculate the integral: $$ \int \int \int _{\Omega}\sqrt{x^2+z^2}\ dx\ dy\ dz $$ Where $\Omega$ it's the region bounded by $y\ =\ x^2 +z^2$ and the plane $y\ =\ 4$

I've performed a change of coordinates to cilindrical considering $y$ as my height so $$\int \int \int _{\Omega}\sqrt{x^2+z^2}\ dx\ dy\ dz\ =\ \int\int\int r^2\ dy\ dr\ d\theta$$

Now i'm having a hard time on setting the itegration limits I've tried with

$r^2 \leqslant y \leqslant 4$, $0\leqslant r\leqslant 2$, $0\leqslant \theta\ \leqslant 2\pi$

Are this well set?

Answer 1

By cylindrical coordinates

• $x=r\cos \theta$
• $z=r\sin \theta$
• $y=y$

we have

$$\int_0^{2\pi} d\theta \int_0^4 dy \int _0^\sqrt yy\,r\ dr$$

Answer 2

Yes. And therefore your integral is$$\int_0^{2\pi}\int_0^2\int_{r^2}^4f(r\cos\theta,y,r\sin\theta)r\,\mathrm dy\,\mathrm dr\,\mathrm d\theta,$$where $f(x,y,z)=\sqrt{x^2+z^2}=r$.