- #1
Telemachus
- 820
- 30
Hi there, I have to compute the surface area for
V:\{ -2(x+y)\leq{}z\leq{}4-x^2-y^2 \}
I have a problem on finding the surface area for the paraboloid limited by the plane. I've parametrized the plane in polar coordinates, I thought it would be easier this way, but also tried in cartesian coordinates with the same result. The problem I have is to set the limits of integration on such a way that determines the area of the paraboloid limited by the plane.
So the paraboloid is parametrized by:
x=r\sin \theta,y=r \cos \theta,z=4-r^2
Then T_r=( \sin \theta, \cos \theta,-2r);T_{\theta}=(r\cos\theta,-r\sin\theta,0)
T_r \times T_{\theta}=(-2r^2\sin^2 \theta,-2r^2\cos^2\theta,-r)
||T_r \times T_{\theta}||=\sqrt[ ]{4r^4+r^2}
Now the surface area is determined by: \displaystyle\int_{D} ||T_r \times T_{\theta}||drd\theta
Now D is the region inside the disc determined by the intersection of the surfaces. So D:
-2x-2y=4-x^2-y^2\rightarrow (x-1)^2+(y-1)^2=6
And this is my region of integration for my paraboloid. Now it doesn't seem so easy to express the region for the parametrization I choose.
So I'm trying to solve this in cartesian coordinates, this is the integral for the surface in cartesian coord:
\displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx
But this integral isn't so easy to solve. I tried to go from here to cylindrical coordinates, using the substitution x=1+r\cos\theta,y=1+r\sin\theta that helps a bit with the limits of integration.
Then I get: \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx=\displaystyle\int_{0}^{\sqrt[ ]{6}}\int_{0}^{2\pi}r\sqrt[ ]{16+8r\cos\theta+8r\sin\theta+4r^2}d\theta dr
Thats the best expression I get, but still too complicated to integrate by hand.
V:\{ -2(x+y)\leq{}z\leq{}4-x^2-y^2 \}
I have a problem on finding the surface area for the paraboloid limited by the plane. I've parametrized the plane in polar coordinates, I thought it would be easier this way, but also tried in cartesian coordinates with the same result. The problem I have is to set the limits of integration on such a way that determines the area of the paraboloid limited by the plane.
So the paraboloid is parametrized by:
x=r\sin \theta,y=r \cos \theta,z=4-r^2
Then T_r=( \sin \theta, \cos \theta,-2r);T_{\theta}=(r\cos\theta,-r\sin\theta,0)
T_r \times T_{\theta}=(-2r^2\sin^2 \theta,-2r^2\cos^2\theta,-r)
||T_r \times T_{\theta}||=\sqrt[ ]{4r^4+r^2}
Now the surface area is determined by: \displaystyle\int_{D} ||T_r \times T_{\theta}||drd\theta
Now D is the region inside the disc determined by the intersection of the surfaces. So D:
-2x-2y=4-x^2-y^2\rightarrow (x-1)^2+(y-1)^2=6
And this is my region of integration for my paraboloid. Now it doesn't seem so easy to express the region for the parametrization I choose.
So I'm trying to solve this in cartesian coordinates, this is the integral for the surface in cartesian coord:
\displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx
But this integral isn't so easy to solve. I tried to go from here to cylindrical coordinates, using the substitution x=1+r\cos\theta,y=1+r\sin\theta that helps a bit with the limits of integration.
Then I get: \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx=\displaystyle\int_{0}^{\sqrt[ ]{6}}\int_{0}^{2\pi}r\sqrt[ ]{16+8r\cos\theta+8r\sin\theta+4r^2}d\theta dr
Thats the best expression I get, but still too complicated to integrate by hand.
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