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In this question, is it right to let $x^2+z^2 = r^2$ and $x=r\cos \theta$, $z=r\sin \theta$?

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    Hint: by careful application of the divergence theorem. – 2017-02-22

2 Answers 2

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Let's introduce the following cylindrical coordinates $$x=r\cos\theta,\ y=y, \ z=r\sin\theta.$$

Then, indeed $$r=\sqrt{x^2+z^2}.$$

The Jacobian of the substitution is $r$.

So after our integral transformation we have

$$\iiint_Nr^2\ dydrd\theta.$$

The following figure helps identifying $N$:

enter image description here

Finally, we have

$$\int_0^{2\pi}\left[\int_0^2\ r^2\left[\int_{r^2}^41\ dy\right] dr\right]\ d\theta.$$

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Yes, your substitutions are correct because the projection of the paraboloid on the $x, z$ axis is a circle for which it is very convenient to use cylindrical coordinates. Your integral should look something like $$\iint_D \int_{x^2+y^2}^4\sqrt{x^2+y^2} \, dy,dA=\int_0^2 \int_0^{2\pi} \int_{r^2}^4r\sqrt{r^2} \, dy,d\vartheta\,dr$$ Which is an easily computable integral.