# Thread: finding the limits of double integral

1. ## finding the limits of double integral

Im not sure how to find the limits of this

Set up an integral to find the volume of the solid below the graph of f(x,y) = 49 - x2 - y2 and above the plane z = 9.

i think the equation becomes x^2+y^2=40 and the x limits are -sqrt(40) to sqrt(40) but im not sure. could someone please explain to me how i get the limits.

thanks!

2. Originally Posted by acosta0809
Im not sure how to find the limits of this

Set up an integral to find the volume of the solid below the graph of f(x,y) = 49 - x2 - y2 and above the plane z = 9.

i think the equation becomes x^2+y^2=40 and the x limits are -sqrt(40) to sqrt(40) but im not sure. could someone please explain to me how i get the limits.

thanks!
Are you allowed to use polar coordinates? If so, that would simplify the "complexity" of the integral.

3. we havent gotten to polar yet, not till next week..

4. Originally Posted by acosta0809
we havent gotten to polar yet, not till next week..
In that case then, we can say that $-\sqrt{40}\leq x\leq \sqrt{40}$ and $-\sqrt{40-x^2}\leq y\leq \sqrt{40-x^2}$

So we can write the integral as $\int_{-\sqrt{40}}^{\sqrt{40}}\int_{-\sqrt{40-x^2}}^{\sqrt{40-x^2}}49-x^2-y^2\,dy\,dx=4\int_0^{\sqrt{40}}\int_0^{\sqrt{40-x^2}}49-x^2-y^2\,dy\,dx$.

5. sorry to ask... its probably simple but why do the bottom limits change if you pull that four out?

6. Originally Posted by acosta0809
sorry to ask... its probably simple but why do the bottom limits change if you pull that four out?
That happens because of the symmetry of the circle. We can then take half of the original interval as our limits for integration, but to compensate, we need to multiply that result by two to get the other section.

The reason why its 4 instead of 2 is because I cut both x and y intervals in half, so I needed to multiply by 2*2=4.

7. Thank you!