1

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???

Muhammad Khalifa
  • 259
  • Duplicate or specific case of another one. See for generalized case: http://math.stackexchange.com/questions/150251/calculate-the-volume-between-z-x2y2-and-z-2ax2by?rq=1 – 007resu Dec 26 '12 at 11:07
  • 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? – Muhammad Khalifa Dec 26 '12 at 11:17
  • @MuhammadKhalifaTranCer: No. It is not true. – Mikasa Dec 26 '12 at 11:18
  • @BabakSorouh:could you tell me how I can find a relation between r and theta ? – Muhammad Khalifa Dec 26 '12 at 11:26
  • 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. – Mikasa Dec 26 '12 at 11:33
  • @user1710036: it ranges from $r^2$ to $3-2rsin(theta)$ – Muhammad Khalifa Dec 26 '12 at 11:36
  • @BabakSorouh no not yet someone told me i need to translate the axes at (0,-1) but I'm not sure – Muhammad Khalifa Dec 27 '12 at 19:40

2 Answers2

3

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.

Mikasa
  • 67,374
0

$$\int_{-2}^{2}\int_{-1- \sqrt{4-x^2}}^{-1+ \sqrt{4-x^2}}\int_{x^2+y^2}^{3-2y}dxdydz$$

zkutch
  • 13,410