1. ## Multiple Integrals

I've done this problem, however not in the way suggested (which I assume is much quicker and easier!)

By evaluating an appropriate double integral, ﬁnd the volume of the wedge lying between the planes $\displaystyle z=px$ and $\displaystyle z = qx$ ($\displaystyle p > q > 0$) and the cylinder $\displaystyle x^2 + y^2 = 2ax$
(where $\displaystyle a > 0$)

I did it by switching to cylindrical polars and then doing the triple integral as I couldnt see a method using a double integral to get the answer $\displaystyle V=\pi (p-q)a^3$

Using the integral

$\displaystyle \int_{\phi=-\pi/2}^{\pi/2} \int_{r=0}^{2a\cos \phi} \int_{z=qr \cos \phi}^{pr \cos \phi} r dz dr d \phi$

any help with doing this with a better method?

2. so we also have that $\displaystyle z \geq 0.$ first find the volume lying between each plane and the cylinder and then subtract the result. the volume lying between the plane $\displaystyle z=px$ and the cylinder is:

$\displaystyle I(p)=\int \int_R px \ dy dx,$ where $\displaystyle R: \ x^2 + y^2 \leq 2ax, \ 0 \leq x \leq 2a,$ which in polar coordinates becomes: $\displaystyle I(p)=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_0^{2a \cos \theta} pr^2 \cos \theta \ d r d \theta=\pi p a^3.$ therefore: $\displaystyle I(q)=\pi q a^3,$ and thus:

$\displaystyle V=I(p)-I(q)=\pi(p-q)a^3. \ \Box$