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Find the volume V of the solid that: lies under the paraboloid $$z=4−x^2−y^2$$ and above the xy-plane. Also, what will change during your process if the solid lies inside and outside the cylinder given by $x^2+y^2≤1$ and $x^2+y^2≥1$, respectively? Set-up the integrals only in these two cases and sketch the corresponding regions.

So, I solved the initial question by integrating twice using polar coordinates, but the second half of this question has me confused. I'm not sure why anything changes if you give a function and state that the solid exists on both sides of it. Wouldn't the function not serve as a bound for the solid?

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  • $\begingroup$ I think the question wants you to do it for each one seperatly. So find the volume outside the cylinder as one question and find the volume inside as another question. $\endgroup$ – lordoftheshadows Dec 1 '16 at 16:54
  • $\begingroup$ Sounds like you don't have a problem setting up the integrals. Inside or outside the the cylinder sets the limits for $r.$ Inside the cylinder you are bound by $r=1.$ On both side of the cylinder you are bound by $r =2.$ Just the region outside the cylinder is $1<r<2$ $\endgroup$ – Doug M Dec 1 '16 at 16:59
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  • For the first case, the projection of $z=4-x^2-y^2$ in the $xy$ plane is a disc with radius $2$, so: $$ V = \iint_{D=\{(r,\theta)|0\le\theta\le 2\pi, \color{red}{0\le r\le 2}\}} (4-r^2)rdrd\theta $$
  • For the second case, you are inside the cylinder with radius $1$, which is inside the disc with radius $2$, so $$ V = \iint_{D=\{(r,\theta)|0\le\theta\le 2\pi, \color{red}{0\le r\le 1}\}} (4-r^2)rdrd\theta $$
  • For the third case, you are outside the cylinder with radius $1$, and inside the disc with radius $2$, so $$ V = \iint_{D=\{(r,\theta)|0\le\theta\le 2\pi, \color{red}{1\le r\le 2}\}} (4-r^2)rdrd\theta $$
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  • $\begingroup$ Much appreciated, that helps a lot. $\endgroup$ – D. Haines Dec 1 '16 at 17:05

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