# Surface Integral of a sphere (or part there of :P!)

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Hi all,

would anyone be able to give me a little bit of help with this question.

for part (a) i did this....

our sphere is given by $x^2 + y^2 +z^2 = 25$

putting in z = 3 to find out part of the sphere gives:

$x^2 + y^2 = 16$

so we change to the parametric representation of the part of the sphere we are interested in sphere which is

$\vec{r}(\theta,\psi) = (4sin\psi cos\theta) \vec{i} + (4sin\psi sin\theta) \vec{j} +(4cos\psi) \vec{k}$

then

$\vec{r_\theta}(\theta,\psi) = (-4sin\psi sin\theta) \vec{i} + (4sin\psi cos\theta) \vec{j}$

$\vec{r_\psi}(\theta,\psi) = (4cos\psi cos\theta) \vec{i} + (4cos\psi sin\theta) \vec{j} +(-4sin\psi) \vec{k}$

so we get that

$\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi) = -16sin^2\psi cos\theta)\vec{i} - 16sin^2\psi sin\theta \vec{j} -16sin\psi cos\psi \vec{k}$

our unit normal vector is given by:

$\vec{\hat{n}} = -\frac{\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|} = \frac{16sin^2\psi cos\theta)\vec{i} + 16sin^2\psi sin\theta \vec{j} + 16sin\psi cos\psi \vec{k}}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|}$

does this look like i am heading in the right direction for part (a)?

cheers!

-Sarah

nm, got it!

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o lol i just got completely lost after the vector field stuff, i was hoping u just wanted how to get the surface area of some part of a sphere, i thought i cud help..:'(

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Let's begin with the Divergence Theorem

$\nabla \cdot \vec{F}=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}=2z$

Now the easiest way to calculate the volume of revolution is to take as volume element a thin slab with circular cross section and thickness $dz$.

The area of the cross section is $\pi(r^2-z^2)$ and the resulting integral is

$\int^5_3 \pi(25-z^2) 2 z dz=128 \pi$

Now for the surface integral:

For the planar part the direction of the normal is $-\vec{k}$. Thus

$\int \int (xz \vec{i}+yz \vec{j} +\vec{k})\cdot (-\vec{k}) d\sigma=-\int \int d\sigma=-16 \pi$

For the spherical part we have to transform both the coordinates and unit vectors of $\vec{F}$ to the spherical coordinates

$\vec{F}=r\sin{\theta}\cos{\phi}r\cos{\theta} (\vec{F}_r \sin{\theta}\cos{\phi} +\vec{F}_\theta \cos{\theta}\cos{\phi}-\vec{F}_\phi \sin{\phi}$

$+ r\sin{\theta}\sin{\phi}r\cos{\theta} (\vec{F}_r \sin{\theta}\sin{\phi} +\vec{F}_\theta \cos{\theta}\sin{\phi}+\vec{F}_\phi \cos{\phi}$

$+(\vec{F}_r \cos{\theta}-\vec{F}_\theta \sin{\theta})$

Now the surface element is the regular $r^2\sin{\theta}drd\theta d\phi$ directed along $\vec{F}_r$

After taking the dot product between the surface element and $\vec{F}$ only terms with $\vec{F}_r$ will survive. We have the following double integral (r=const)

$\int_0^{2\pi}d\phi \int_0^{\arccos{3/5}} d \theta r^2 \sin{\theta} (r^2 \sin{^2 \theta} \cos{\theta} \cos{^2\phi}+r^2 \sin{^2 \theta} \cos{\theta} \sin{^2\phi}+\cos{\theta})=$

$=\int_0^{2\pi}d\phi \int_0^{\arccos{3/5}} d \theta \left(r^4 \sin{^3 \theta}\cos{\theta} + r^2 \sin{\theta}\cos{\theta} \right)=144 \pi$

Summing the contributions of the planar and spherical parts, we have $-16 \pi + 144 \pi=128 \pi$

The divergence theorem works!!!

Regarding your solution : the direction of the normal to the spherical part when expressed through i,j,k vectors is indeed $\vec{r}(\theta,\phi)$ (6th line from above in your answer). There is no need to calculate the cross-product of $\vec{r}_\theta \times \vec{r}_\phi$ - this will only bring us again to the same radius vector. In order to continue you have to multiply the normal vector by the surface element of the spherical part.

If you have a questions, I'd be happy to answer.

Good luck!

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