# Thread: surface area and integration

1. ## surface area and integration

Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.

Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} )$

I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} ||$ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi$

Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} ||$ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2}$ but am stuck on how to use the hint?

2. Originally Posted by nmatthies1
Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.

Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} )$

I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} ||$ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi$

Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} ||$ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2}$ but am stuck on how to use the hint?
Seems you want to calculate

$\displaystyle \int || r_{\theta} \times r_{\phi} || \; d\theta =\int 2\sin \theta \sqrt{\sin^2 \theta + 4\cos^2 \theta} \; d\theta=-2\int \sqrt{1 + 3\cos^2 \theta} \; d \cos \theta$$\displaystyle = -2\int \sqrt{1 + 3u^2} \; du$ (Let $\displaystyle u=\cos \theta$)

Then use the hint to proceed ....

3. Originally Posted by nmatthies1
Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.
This does NOT give an ellipsoid. $\displaystyle \vec{r}(\theta, \phi)= (2cos\phi sin\theta, 2 sin\phi sin\theta, cos\theta)$ gives the ellipsoid $\displaystyle \frac{x^2}{4}+ \frac{y^2}{4}+ z^2= 1$.

Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} )$ .

I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} ||$ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi$

Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} ||$ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2}$ but am stuck on how to use the hint?
In the form I have given, $\displaystyle \phi$ is the "longitude" and ranges from 0 to $\displaystyle 2\pi$, and $\displaystyle \theta$ is the "co-latitude" and ranges from 0 to $\displaystyle \pi$.