Question

Evaluate the integral, where E is the solid in the first octant that lies beneath the paraboloid z = 1 - x2 - y2. Use cylindrical coordinates. triple integral (4(x^3+xy^3)) dV

Answer 1

z is going to range from:
\[0\le z \le 1-r^2\].
Then let z=0 Solving you get:
\[x^2+y^2=1\].
Which implies:
\[0 \le r \le 1\].
And its in the first octant so:
\[0 \le \theta \le \frac{\pi}{2}\].
Just don't forget when you do dV to make it: \[r dz dr d \theta\]

Answer 2

Then you need to convert your integrand:
\[4(x^3+xy^3)=4(r^3\cos^3(\theta)+r^4\cos(\theta)\sin^3(\theta))\]

Answer 3

So you should have:
\[4\int\limits_{0}^{\frac{\pi}{2}} \int\limits_{0}^{1} \int\limits_{0}^{1-r^2}(r^3\cos^3(\theta)+r^4\cos(\theta)\sin^3(\theta))rdzdrd \theta\].
Not pretty but should be doable :P