surface area and integration

**nmatthies1** · September 10th 2009, 06:40 AM

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

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

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

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

**luobo** · September 10th 2009, 03:52 PM

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 $r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$ .

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

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

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

Seems you want to calculate

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

Then use the hint to proceed ....

**HallsofIvy** · September 11th 2009, 07:31 AM

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 $r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$ .

This does NOT give an ellipsoid. $\vec{r}(\theta, \phi)= (2cos\phi sin\theta, 2 sin\phi sin\theta, cos\theta)$ gives the ellipsoid $\frac{x^2}{4}+ \frac{y^2}{4}+ z^2= 1$ .

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

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

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

In the form I have given, $\phi$ is the "longitude" and ranges from 0 to $2\pi$ , and $\theta$ is the "co-latitude" and ranges from 0 to $\pi$ .

Thread: surface area and integration

LinkBack

Thread Tools

Display

surface area and integration

Similar Math Help Forum Discussions

Very Quick Question Regarding Surface Area of Revolution (Integration)

Surface area integration (flux integral)

Calculate the surface area of the surface

Volume, Surface Area, and Lateral Surface Area

[SOLVED] [SOLVED] Surface Area using integration

Search Tags