# Volume under curve z=1-x^2-2y^2

1. Apr 12, 2012

### brandy

1. The problem statement, all variables and given/known data
z=1-x^2-2y^2
find volume under curve
bounded by the xy plane.
is the answer sheet wrong? (see below)
why am i struggling so much with this??!?!?! how do i do it?

2. Relevant equations
according to other answer sheet, it is pi/sqrt 2

3. The attempt at a solution
i did this, its bound by z=0 and it should be the same in all the upper quadrants so i tried too work out one quadrant and went from there.
4*integral(0 to 0.25) (integral(0 to 1)(1-x^2-2y^2) dx)dy =0.625
although im wondering if it should be
4*integral(0 to 1) (integral(0 to 0.25sqrt(1-x^2)(1-x^2-2y^2) dy)dx =0.564504

2. Apr 12, 2012

### Staff: Mentor

The limits on the inner integral aren't right.
This setup looks right, but I haven't carried out the integration, so can't confirm that your answer is correct.

3. Apr 12, 2012

### NewtonianAlch

It would help after 150 posts if people made the effort to put at least some parts of it into latex.

4. Apr 12, 2012

### Staff: Mentor

I agree.

brandy, here is your second integral using LaTeX. Right-click the integral to see my LaTeX script.

$$4\int_{x = 0}^1\int_{y = 0}^{.25\sqrt{1 - x^2}}1 - x^2 - y^2~dy~dx$$