volume under curve z=1-x^2-2y^2

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

Mark44

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.

NewtonianAlch

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

Mark44

I agree.

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

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