# How do I find the limits of the outermost integral in a triple integral representing a 3D solid? So far what I've got for solving this problem is:

$$\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \int_{\sqrt{x^2+y^2}}^{3} dz dx$$

And I think that's correct, but I'm not sure.

How do I figure out the limits of the outermost integral (which will be with respect to $$x$$)? I can't use $$\sqrt{9-x^2}$$ because the limits of the outermost integral must be constants. But that equation is the only information I have about $$x$$, isn't it? Do I just set the limits to be from $$0$$ to $$1$$?

Any help is appreciated.

Remember the mass is given by \begin{align*} M = \iiint_{D}m(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z \end{align*}
In the present case, $$D = \{(x,y,z)\in\textbf{R}^{3} \mid \sqrt{x^{2}+y^{2}}\leq z \leq 3\}$$ and $$m(x,y,z) = z$$. Thus we have to solve the following integral \begin{align*} \int_{-3}^{3}\int_{-\sqrt{9-x^{2}}}^{+\sqrt{9-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{3}z\mathrm{d}z\mathrm{d}y\mathrm{d}x \end{align*}
• Thank you! Indeed, I could take it from there, but how do you know that the upper and lower limits on the outermost variable are $3$ and $-3$? Also, could you have switched the two outer integrals (like how I did: the integral with respect to $y$ is the outermost integral, while in yours is the middle integral)? Or can you not switch them in this case? – James Ronald Sep 14 at 19:30
• I like to reason as it follows: first, you determine the length of the line segment whose end points are $\sqrt{x^{2}+y^{2}}$ to 3. Then you determine the area determined by adding up all these line segments' lenghts letting $y$ varying from $-\sqrt{9-x^{2}}$ to $\sqrt{9-x^{2}}$. Finally, you determine the volume corresponding to the sum of all these areas by letting $x$ varying from -3 to 3. Since you are interested in finding the mass, we must multiply the integrand by z (in this case). But you can switch the order of integration taking the care to interpret correctly what is happening. – user1337 Sep 14 at 19:40