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.
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.
