Question

Use a triple integral to find the volume of the given solid:

The solid enclosed by the cylinder x^2 + y^2 = 9 and the planes y + z = 5 and z=1.
Please explain step by step.

Answer 1

T_T. Calc 3. brutal man. use circular coordinates i think? turn x^2+y^2 = 9 into r^2 = 9.
r will be from 0 to 3. Thats one bound. z = 5-y. another bound is z goes from 1 to (5-y). thats a start, Im having a hard time remembering the variables you use for this type.

Answer 2

the last variable u use is theta i belive. from 0 to 2pi. turn y into rsin(theta).

The three will be 0->3, 0->2pi, 1->(5-y) S(r*dz*d(theta)*dr). Something similiar. This will not work out...at least I dont think it will. sorry

Answer 3

first sketch to have a better view on the problem so we'll have:


as we can see the upper and lower limits for our integral will be
\[-3\le x \le3\]
\[-\sqrt{9-x^2} \le y \le \sqrt{9-x^2}\]
\[1 \le z \le 5-y\]
so our triple integral will become:
\[\int\limits_{-3}^{3}\int\limits_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int\limits_{1}^{5-y}dV\]