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

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
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Mentor
 Quote by 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
The limits on the inner integral aren't right.
 Quote by brandy 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
This setup looks right, but I haven't carried out the integration, so can't confirm that your answer is correct.
 It would help after 150 posts if people made the effort to put at least some parts of it into latex.

Mentor

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

I agree.

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

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

 Tags domain, double integrals, integration, three space, volume

 Similar discussions for: volume under curve z=1-x^2-2y^2 Thread Forum Replies Biology 8 Calculus & Beyond Homework 3 Calculus & Beyond Homework 4 Calculus & Beyond Homework 2 Introductory Physics Homework 7