Triple Integrals with Cylindrical Coordinates

dancingmonkey
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Homework Statement


Evaluate the integral, where E is the solid in the first octant that lies beneath the paraboloid z = 9 - x2 - y2.

∫∫∫(2(x^3+xy^2))dV

Homework Equations



x=rcosθ
y=rsinθ
x^2+y^2=r^2

The Attempt at a Solution



θ=0 to 2π, r=0 to 3, z=0 to (9-r^2)

2(x^3+xy^2)=2x(x^2+y^2)=2rcos(θ)(r^2)

∫0 to 2π ∫0 to 3 ∫0 to (9-r^2) (2rcos(θ)r^2)rdzdrdθ

I was wondering if my bounds were correct. And when I solved the integral I keep getting an answer of 0, which is incorrect. Can someone please help me with this problem?
 
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If you're integrating over the first octant, then theta doesn't go from 0 to 2pi. Other than that, the rest is ok.
 
Thank you so much! That was the problem. I missed the part where it said the first octant.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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