Find the volume of the solid in the first octant bounded by the cylinder $z=9-y^2 $ and the plane $x=2$

Can I solve this problem using triple integrals in the following way

$$\int_0^2\int_0^{\sqrt{9-z}}\int_0^{9-y^2}1 \, dzdydx$$

I'm currently studying double integrals in my course but I'm not entirely sure how to attack the problem that way. Doing a bit of research I found a problem about a solid prism with similar bounds. I was wondering if I could solve the problem with triple integrals and if so would it be a better option than with double?

I'm not looking for a solution just to let you know. Thanks


1 Answer 1


To evaluate the volume of this solid, a triple iterated integral is fine. Note that the solid is given by $$\{(x,y,z)\in\mathbb{R}^3\;:\; 0\leq x\leq 2,\; 0\leq y,\; 0\leq z\leq 9-y^2\}.$$ So your upper integration limit for $y$ is not correct (and according to the order of integration, in any case it should not depend on $z$). What is the right upper limit for $y$?

  • $\begingroup$ Right! if I let $z=0$ I get $y=\pm 3$. Since $y \ge 0$ that gives the bounds as $$0 \le y \le 3$$ $\endgroup$ Oct 26, 2017 at 13:43
  • $\begingroup$ @PatrickMoloney Correct! $\endgroup$
    – Robert Z
    Oct 26, 2017 at 13:46

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