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Let $$V = \{(x, y, z): x^2 + y^2 ≤ 4 , 0 ≤ z ≤ 4\}$$ be a cylinder and let $P$ be the plane through $(4, 0, 2), (0, 4, 2)$ and $(−4, −4, 4)$. Compute the volume of $C$ below the plane $P$.

I'm having trouble trying to start this question. I believe you use spherical coordinates but then again I'm not too sure. Please help. Thanks.

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  • $\begingroup$ Have you heard of cylindrical coordinates ? This problem is doable but overly complex in spherical coordinates, and trivial in cylindrical ones. $\endgroup$ – Yves Daoust May 6 '15 at 10:23
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Since it is a cylinder and $z$ values are given, you should use cylindrical coordinates: $0\leq r\leq 2, 0\leq \theta \leq 2\pi$. $z$ values should be from $0$ to the plane.

So you need to find the plane in the form of $z=ax+by+c$, and set that as your upper limit of $z$ value.

The plane passes through two points, so the normal direction will be the cross product of two vectors $(-4,4,0)$ and $(-4,-8,2)$, which are obtained by subtracting pairs of two points.

After you find the normal direction $(a,b,c)$, you can set your plane as $ax+by+cz=d$. Plugging one point into it will give you $d$.

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  • $\begingroup$ How would you go about finding the equation of the plane and what is the integrating function? $\endgroup$ – Jim May 6 '15 at 10:05
  • $\begingroup$ @Jim: You can either do a triple integral with the limits as mentioned in the answer, and integrand $1$, or a double integral with the plane function in it. I will edit how to find the plane. $\endgroup$ – KittyL May 6 '15 at 10:17
  • $\begingroup$ I worked out the equation of the plane to be z=(16-x-y)/6, so do I do triple integrals for 0 to (16-x-y)/6 then 0 to 2 pie and then 0 to 2? $\endgroup$ – Jim May 6 '15 at 10:27
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    $\begingroup$ @Jim: Yes. Remember for cylindrical coordinates, you should have $rdrd\theta$ for the change of variables. $\endgroup$ – KittyL May 6 '15 at 10:39

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