Setting up Double Intergrals to Calculate Volume

1. The problem statement, all variables and given/known data
Set up the double integral in rectangular coordinates for calculating the volume.
The solid is created by the equation



2. Relevant equations
Fubini's Thereom ∫_R∫f(x,y)dA = ∫_a^b∫f(x,y)dydx


3. The attempt at a solution
The shape created by Z = 30 - x² - y² is 3d parabola facing downards.
Since it is bounded by Z = 5, the base shape is a circle at origin with radius 5 since
5 = 30 - x² - y² is equal to x² + y² = 25
As a result x is bounded by -5 and 5 while y is bounded by -√(25-x^2) and √(25-x^2)

As a result to set up the double integral
∫_-5⁵∫_√(25-x^2)^{√(25-x^2)}30-x^{^2}-y^{^2}dydx

Are the integral boundaries in correct?

 

Thanks for the response! However, that is the answer I entered into Webwork but it states it is incorrect.

 

Okay I managed to figure it out after reading: http://mathinsight.org/double_integral_volume

My boundaries were correct but the equation I was integrating was not.
It explained that the integral equation is not f(x,y) but f(x,y) - g(x,y), in my case g(x,y) being the "other surface" or z = 5

As a result I subtracted 5 from f(x,y) which was 30 - x^2 - y^2 and resulted in a final answer of
∫_-5⁵∫_-√(25-x^2)^{√(25-x^2)} 25-x^2-y^2dydx