- #1

hbomb

- 58

- 0

S is the solid bounded above by the sphere [tex]x^2 + y^2 + z^2 = 8[/tex]

and below by the paraboloid [tex]x^2 + y^2 = 2z[/tex]. Solve this using spherical coordinates

The correct answer is [tex]\displaystyle{\frac{4\Pi}{3}}[/tex][tex](8\sqrt{2}-7)[/tex]

Here is the solution that I achieved...

[tex]V=\int\int\int dV[/tex]

After drawing the 3d graph, I decided to just take the volume of one octant and multiply that by 4 (since the graph is only symmetrical about the z-axis and lies above the xy plane.

[tex]4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}\thet }\;\rho^2\;\sin\phi\;d\rho\;d\theta\;d\phi[/tex]

[tex]4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\thet }\;{\frac{\rho^3}{3}}\;\sin\phi\;\mid_{0}^{\2\sqrt{2}}\;d\theta\;d\phi[/tex]

[tex]\frac{64\sqrt{2}}{3}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\thet }\;\sin\phi\;d\theta\;d\phi[/tex]

[tex]\frac{64\sqrt{2}}{3}\int_{0}^{\frac{\pi}{2}}\sin\phi\;\theta\;\mid_{0}^{\frac{\Pi}{2}}\;d\phi[/tex]

[tex]\frac{32\sqrt{2}\Pi}{3}\int_{0}^{\frac{\pi}{2}}\sin\phi\;d\phi[/tex]

[tex]\frac{-32\sqrt{2}\Pi}{3}\cos\phi\;\mid_{0}^{\frac{\Pi}{2}}[/tex]

[tex]\frac{4\Pi}{3}\(8\sqrt{2}[/tex]

Where is the -7?

I don't see any place I went wrong at.