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I am having trouble just setting up the integrals for this problem.

Find the volume of the solid bounded by $x^2 + y^2 = 1, z = 0$, $z = 6$, $y\geq 1/2$.

a) Use integration with Cartesian coordinates.

b) Use integration with Cylindrical coordinates.

c) Use integration with Spherical coordinates. (Hint: Use two triple integrals and tangent inverse)

The solid should look like this.

Assuming I used Mathematica correctly, I get this.

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I will only give an answer for the Cartesian method, to help you on your way.

For both Cartesian and Cylindrical Polar methods, you need to integrate between $z = 0$ to $z = 6$ the cross section $x^2 + y^2 = 1, y≥\frac{1}{2}$

In Cartesian, we clearly have $\frac{1}{2} ≤ y ≤1$ and then $-\sqrt{1-y^2}≤x≤\sqrt{1-y^2}$ from our equation of the circle. So our integral is thus: $$\int_{z=0}^6\int_{y=\frac{1}{2}}^1\int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ I recommend making the substitution $y=\sin\theta$ to solve the integrand with the square root.

The Cylindrical Polar method is very similar: just remember your Jacobian.

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  • $\begingroup$ Thanks! I got the answers for both parts a and b, but I'm still having a tough time with part c. I'm not sure how to even start or why I would even need a tangent inverse. $\endgroup$ – bobtran12 Apr 19 '15 at 7:26
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    $\begingroup$ Take a look at this question and its answers math.stackexchange.com/questions/614969/… $\endgroup$ – elDin0 Apr 19 '15 at 7:39

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