# Solving for the Volume of a Solid Using Double Integrals

1. Mar 9, 2014

### Temp0

1. The problem statement, all variables and given/known data
Find the volume of the solid bounded above by the surface z = x^2 + y^2 and below by the
triangular region in the xy-plane enclosed by the lines x = 0 , y = x , and x + y = 8.

2. Relevant equations
V = ∫∫ Height
Base

3. The attempt at a solution
I first found height, because height = z (upper) - z (lower) = (x^2 + y^2) - 0 = x^2 + y^2

Afterwards, I began solving for the base. I know the base is enclosed by the curves y = x,
y = x - 8, and x = 0. I proceed assuming the region is y-simple, giving me the boundaries:
Base { x ≤ y ≤ 8-x
and 0 ≤ x ≤ 4 }
This sets up the double integral for me, which turns out to be:
4 8-x
∫ { ∫ (x^2 + y^2) dy } dx
0 x

Solving this integral gives me that volume is 1024/3.
Could someone look through my work and see if I made any errors? I'm a beginner at setting up these equations, so I suspect my integral to have a mistake. Thank you in advance.

2. Mar 9, 2014

### Zondrina

Hmm I'm getting $\frac{2048}{3}$. Double your answer.

The way I managed to get that was by starting with $x = y$ and $x = 8 - y$.

Then $x = 0$ implies that $y = 0$ and $y = 8$.

3. Mar 10, 2014