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
Best Answer
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 $y$?