Calculate the mass of the surface of a sphere, given density is equal to the distance from the xy-plane?

Question

Question: $ \text{Find the mass of the surface } x^2 +y^2+z^2 = 16 \text{ where the density at a point is the distance to the xy-plane.} $

Distance from the xy plane can be given in terms of x and y is equal to z. So I solve for z and get $ z = \sqrt{16 - y^2 - x^2} $ I know that both y and x range from [0,4] to get me 1/8 of the sphere.

So the problem then becomes the following:

$ 8 \int_{0}^{4} \int_{0}^{4} \sqrt{16 - y^2 - x^2} dy dx$

But obviously that's a mess which makes me think I should be using spherical coordinates but I'm struggling to figure out how to work the density into that equation.

Answer 1

The spherical integral for the surface mass is

$$I = \int_S |z|r^2\sin\theta d\theta d\phi$$

where $|z|$ is the density. Plug in the radius $r=4$ and the distance

$$z=r\cos\theta,$$

and then carry out the integration

$$I=64 \int_0^{2\pi} d\phi \int_0^{\pi} \sin\theta|\cos\theta| d\theta = 128\pi$$

Answer 2

First you want to find out the function $\vec{F}$. In this case, it is simply the z-component of the surface: $$\vec{F}=<0,0,z>$$ You are trying to compute the surface integral over a vector field: $$\int \int_S \vec{F} \cdot d\vec{S} = \int\int_{(u,v)\in D} \vec{F}(\vec\Phi) \cdot(\vec\Phi_u\times\vec\Phi_v)\ du\ dv$$ Next, you want to figure out how you want to parameterize the function. Since the surface is a sphere, it would stand to reason that you will want to use spherical coordinates. $$\vec\Phi(\phi,\theta)=<\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi>$$ $$\rho = \sqrt{16} = 4$$ $$0\leq\theta\leq 2\pi $$ $$0\leq\phi\leq \pi $$ So, $$\vec\Phi(\phi,\theta)=<4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi>$$ Take the partial derivatives in terms of $\phi$ and $\theta$. $$\vec\Phi_\phi(\phi,\theta)=<4\cos\phi\cos\theta,4\cos\phi\sin\theta,-4\sin\phi>$$ $$\vec\Phi_\theta(\phi,\theta)=<-4\sin\phi\sin\theta,4\sin\phi\cos\theta,0>$$ Then get the cross product of the two terms to get the normal vector: $$\vec\Phi_\phi(\phi,\theta)\times\vec\Phi_\theta(\phi,\theta) = <16\sin^2\phi\cos\theta, 16\sin^2\phi\sin\theta, 16\sin\phi\cos\phi> $$ And get the function with regards to $\Phi$: $$\vec{F}(\vec\Phi(\phi,\theta)) = <0,0,4\cos\phi> $$ Next you dot product the function and the normal vector: $$\vec{F}(\vec\Phi(\phi,\theta)) \cdot(\vec\Phi_\phi(\phi,\theta)\times\vec\Phi_\theta(\phi,\theta)) = 0+0+64\sin\phi\cos^2\phi$$ Then set up your integral: $$\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} 64\sin\phi\cos^2\phi\ d\phi d\theta $$ $$\frac{128}{3}\int_{\theta=0}^{2\pi}d\theta = \frac{256\pi}{3}$$

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Alternatively, since the surface is closed, you can utilize the Divergence Theorem (Gauss's Theorem) to calculate the mass. $$\int\int\int_E Div\vec{F}\ dV $$ $$Div\vec{F} = \vec{\nabla} \cdot \vec{F} = \left <\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right > \cdot \left <0,0,z \right > = 0+0+1 = 1$$ In this case: $$dV = \rho^2\sin\phi\ d\rho\ d\phi\ d\theta$$ and: $$\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi}\int_{\rho=0}^{4} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta $$ $$=\frac{64}{3}\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} \sin\phi\ d\phi\ d\theta$$ $$=\frac{128}{3}\int_{\theta=0}^{2\pi} d\theta=\frac{256\pi}{3}$$