Set up a double integral in rectangular coordinates for calculating the volume of the solid under the graph of the function f(x,y) = 30-x^{2}-y^{2} and above the plane z = 5.

I ended up with the double integral of 30-x^{2}-y^{2} dydx. where the limits were from (-5 to 5) for the first, and (-sqrt(25-x^2, and sqrt(25-x^2).

It says I am wrong, and I'm kind of lost!

Question

Set up a double integral in rectangular coordinates for calculating the volume of the solid under the graph of the function  f(x,y) = 30-x^{2}-y^{2} and above the plane z = 5. 

I ended up with the double integral of 30-x^{2}-y^{2} dydx. where the limits were from (-5 to 5) for the first, and (-sqrt(25-x^2, and sqrt(25-x^2). 

It says I am wrong, and I'm kind of lost!

majikdusty · Answer

$$\int\limits_{-5}^{5}\int\limits_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}(30-x^2-y^2)dydx$$

This is what I got, and it is considered to be wrong. Is there maybe something I am missing with the rectangular coordinates??