When I try Find the volume of the region $R$ lying below the plane $z = 3-2y$ and above the paraboloid $z = x^2 + y^2$

Solving the 2 equations together yields the cylinder $x^2 + (y+1)^2 = 4 $ How do I get the volume then???

  • $\begingroup$ Duplicate or specific case of another one. See for generalized case: math.stackexchange.com/questions/150251/… $\endgroup$ – 007resu Dec 26 '12 at 11:07
  • $\begingroup$ what i don't understand is that when using polar coordinate to integrate my book deals with the region (which is the circle: x^2 + (y+1)^2 = 4) as if its center was the origin so theta: 0 --> 2PI , r: 0 --> 2 is that true? $\endgroup$ – Muhammad Khalifa Dec 26 '12 at 11:17
  • $\begingroup$ @MuhammadKhalifaTranCer: No. It is not true. $\endgroup$ – mrs Dec 26 '12 at 11:18
  • $\begingroup$ @BabakSorouh:could you tell me how I can find a relation between r and theta ? $\endgroup$ – Muhammad Khalifa Dec 26 '12 at 11:26
  • $\begingroup$ The problem is that the intersection area is not a circle with centered at (0,0). its center is indeed $(0,-1)$ with radius 2 as you noted above. $\endgroup$ – mrs Dec 26 '12 at 11:33

First of all, I draw a plot for $x^2+(y+1)^2=4$ or $r^2+2r\sin(\theta)=3$ which is our integration area on plane $z=0$.

enter image description here

You see that $r$ varies from $r=3$ to $r=-\sin(\theta)+\sqrt{\sin(\theta)^2+3}$ and $\theta$ from $0$ to $pi/2$. As the volume is symmetric so you should double the result.

  • $\begingroup$ Wow! I like the picture (and the answer!) + $\endgroup$ – Namaste Mar 1 '13 at 0:59
  • $\begingroup$ @amWhy: I my self made it! :-) $\endgroup$ – mrs Mar 1 '13 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.