# Mass of Solid, Multivariable Calculus

Calculate the mass of the solid that lies above the surface $z= 0$, below the surface $z=y$, and inside the surface $x^2+y^2 = 4$ with the given density $yz$. I have switched to cylindrical coordinates, and carried out integration -- but, I don't know if my limits are right. My limits are $0\leq \theta ≤ 2π , 0≤z≤r\sin\theta$, and

• If you don't convert to cylindrical coordinates, the limits are $$\int_{-2}^2dy\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dx\int_0^ydz$$ – Arthur Dec 11 '14 at 12:27
• I made a mistake. It's supposed to be $\int_0^2 dy$, since otherwise you get a contribution from the solid on the other side (above $y = z$ and below $z = 0$). – Arthur Dec 11 '14 at 12:33
• Are my limits correct, though ? @Arthur – None Dec 11 '14 at 18:22
• $\theta$ goes from $0$ to $\pi$. Otherwise it's correct. – Arthur Dec 11 '14 at 20:42

Cylindric coordinates with the coordinate cylinder aligned with the cylinder $x^2+y^2=4$ in your question means that you use $\theta,z,r$, where $z$ is simply one of your cartesian coordinates and the other two are defined by

$$r = \sqrt{x^2+y^2} \qquad \sin\theta=\frac yr \qquad \cos\theta=\frac xr$$

So let's see about your definitions. $z\ge0$ you can simply keep. $x^2+y^2\le4$ is simply $r\le2$. So the tricky thing is $z=y$. To make this work, you'll either have to express the range for $\theta$ as a function of $z$ or vice versa. The latter is far simpler, since the second equation above readily translates to $y=r\sin\theta$. So yes, the relevant portion of the cylinder can be described by

$$0\le r\le2 \qquad 0\le z\le r\sin\theta$$

with no condition on $\theta$. Now if you want to use this for integration, you need to be a bit more careful. On the one hand, you want to cover every matching point exactly once, so you want to restrict $\theta$ to go around exactly once. That's the $0\le\theta\le2\pi$ in your question. If you use a pair of inequalities as bounds for a variable, you also have to make sure that these bounds are ordered as expected, because otherwise you might end up integrating negative volume for parts outside your solid. In this specific case you want to ensure $0\le r\sin\theta$ which, since $r\ge0$, restricts $\theta$ to $0\le\theta\le\pi$ just as Arthur wrote in a comment.

So you could formulate your integral as

$$\int_0^2\mathrm dr\; \int_0^\pi r\,\mathrm d\theta\; \int_0^{r\sin\theta}\mathrm dz\; \rho(r,\theta,z)$$

But if your density $\rho$ is really $yz$, then I'm not sure that cylindric coordinates are such a good idea, since you'd have $\sin\theta$ all over the place.