4
$\begingroup$

Consider the paraboloid $(\mathcal{P}): z=x^2+y^2$ and the plane $(\mathcal{Q}): 2x+2y+z=2$. Let $\mathcal{S}$ be the solid region bounded above by $(\mathcal{Q})$ and below by $(\mathcal{P})$. Find the volume of the solid region $\mathcal{S}$.

After sketching the plane will cut the paraboloid and obtain the shadow in the $xy$-plane as a circle of equation $$ R: (x+1)^2+(y+1)^2=4. $$ Now to find the volume, I travel first in the $z$-direction and get $$ V(\mathcal{S})=\iint_{R}\int_{x^2+y^2}^{2-2x-2y}dz dA=\iint_{R}\left(2-2x-2y-x^2-y^2\right)dA. $$ Now to evaluate the double integral over the shadow $R$, we may use cartesian coordinates and try to evaluate it, but the calculation are very hard. Is it possible to find the volume for example using cylindrical or spherical coordinates? How we can do it?

Based on the comment below, $r^2=(x+1)^2+(y+1)^2$, the volume could be $$ V(\mathcal{S})=\int_{0}^{2\pi} \int_{0}^2\left[-r^2+4\right]r dr d\theta. $$ Is it correct in this way to solve it?

$\endgroup$
7
  • 2
    $\begingroup$ Yes- cylindrical coordinates would be ideal in this situation because like you said, the shadow is a circle. You can translate the center of the radius to be at (-1,-1) and do regular cylindrical integration with the origin slightly shifted. $\endgroup$
    – D P
    Apr 3 at 0:05
  • $\begingroup$ @DP ok great, but how we can use the translation of the center here? what would be the limits of integration for r? $\endgroup$
    – Student
    Apr 3 at 13:50
  • $\begingroup$ you can let $r^2=(x+1)^2+(y+1)^2$ and the limits of integration would be from 0 to 2 (because the radius increases from 0 to 2) The angle theta goes around the circle like normal, so from 0 to $2\pi$. The z direction is a little more cumbersome, but because you already evaluated it from $x^2+y^2$ to $2-2x-2y$ all you need to do is find $2-2x-2y-x^2-y^2$ in terms of r, which in this situation happens to be quite convenient. $\endgroup$
    – D P
    Apr 3 at 20:26
  • $\begingroup$ @DP Thank you for your comment. I have edited my question based on your comment, is it correct now? $\endgroup$
    – Student
    Apr 4 at 16:56
  • $\begingroup$ yes- this looks correct. You can solve it now using regular integration techniques. $\endgroup$
    – D P
    Apr 4 at 16:59

0

You must log in to answer this question.