I am having trouble finding the volume between the two surfaces. Well I did the problem and got an answer but im not sure if im approaching this type of problem correctly. So can you take a look at how i approached the problem and tell me if i am on the right path or if it is incorrect. Thank you in advance. Find the volume between the two surfaces z = x^2 + 3y^2 z = 8 - x^2 - y^2 First I set both equations equal to each other to get y =±sqrt(2 - x^2). Finding the volume will require a double integral so this can be my y limits. X cannot exceed sqrt(2) so i assume my x limits are ±sqrt(2) The sum of the two functions is 2y^2 + 8. I believe this is what i should integrate. If not please tell me what i should integrate and why. So i integrate 2y^2 + 8 with respect to y and get (2/3)y^3 + 8y. And inputting the y limits yields (4/3)*(2 - x^2)^(3/2) + 16*(2 - x^2)^(1/2). Now i integrate that with respect to x using the computer and get 18*arcsin(sqrt(2)x/2) - (1/3)x*(x^2 - 5)*sqrt(2 - x^2) + 8x*sqrt(2 - x^2) Putting in limits of ±sqrt(2) gets a final answer of 18*pi.