Multiple Integrals

**thelostchild** · December 19th 2008, 04:21 AM

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 $z=px$ and $z = qx$ ( $p > q > 0$ ) and the cylinder $x^2 + y^2 = 2ax$
(where $a > 0$ )

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 $V=\pi (p-q)a^3$

Using the integral

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

any help with doing this with a better method?

**NonCommAlg** · December 19th 2008, 12:17 PM

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

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

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

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