Volume integral over a bounded region

Question

Class is over now and I am studying for my final and I have a problem with this question on our review sheet. If anyone can help I would appreciate it.

Question: Find the volume of the region in space bounded by $z=1-x^2-y^2$ and on the sides by the planes x=0, y=0 and x+y=1 and z=0.

I first found the boundaries: $0\le z \le 1-x^2-y^2$ , $0\le y \le 1-x$, and $0\le x \le 1$.

Originally I just then computed the integral $\int\int\int dzdydx$ which gave me the answer of 1/3 however I feel as though I should be using polar coordinates for this question but when I tried that I could not seem to figure out anything that made sense. I might be screwing up the integral when I switch over to polar coordinates. Worst part is our teacher hasn't posted any solutions so I am really shooting in the dark. Again if anyone can help I would appreciate it.

Answer 1

If you switch coordinate systems, I would recommend cylindrical coordinates (r, theta, z).

Then z = 1 - r^2 and the dx dy would become r dr d(theta), you would also need to change your limits of integration. theta would be 0 to pi/2, r from 0 to 1/(sin(theta) + cos(theta)).

Solving in Cartesian coordinates: $\\ \int_{x=0}^1\int_{y=0}^{1-x}\int_{z=0}^{1-x^2-y^2}dzdydx = \\ \int_{x=0}^1\int_{y=0}^{1-x} 1 - x^2 - y^2 dydx =\\ \int_{x=0}^1 (1 - x^2)(1 - x) - (1/3)(1 - x)^3 dx =\\ \int_{x=0}^1 (4/3)x^3 - 2x^2 + (2/3) dx = 1/3$

Answer 2

You did the limits right in Cartesian coordinates, but we can, as you pointed, do the triple integral by using the Cylindrical coordinates: $$0\leq z\leq 1-x^2-y^2\longrightarrow 0\leq z\leq 1-r^2 $$ since $x^2+y^2=r^2$. Since $$x+y=1\longrightarrow r(\sin\theta+\cos\theta)=1\\\ z=0\longrightarrow1=x^2+y^2\longrightarrow r=1$$ so the ranges of changing $r$ would be as: $$r|_{0}^{\frac{1}{(\sin\theta+\cos\theta)}}$$ For $\theta$, we have $$\theta|_{0}^{\pi/2}$$