1. ## area of ellipsoid

Can somebody please explain how to do the following question, I am struggling to get started and also don't know why it is asking for an area as opposed to a volume.

1.(i). Calculate the area of the ellipsoid (x^2/4) +(y^2/4)+z^2=1 that lies above the xy plane using the parameterisation:
r(θ,φ)=(2cosφsinθ, 2sinφsinθ, cosθ).

2. $\displaystyle Area = \iint_S \left|\vec r'_\varphi \times \vec r'_\theta\right|d\theta d\varphi$

Surface integral - Wikipedia, the free encyclopedia

The surface in question is the upper half of an ellipsoid (only the positive part of the z axis).

3. See attachment in later post

4. Originally Posted by Calculus26
C.E

See Attachment

double check my Calculations
How on earth did you obtain

$\displaystyle || \vec{r}_{\phi} \times \vec{r}_{\theta}|| = 4(\sin^2 \theta + \sin^2 \theta \cos^2 \theta)?$

5. That's why i said double check my calculations--- i rushed

what did you come up with for comparison sake?

6. Does 4 *sin^4(t) +16sin^2(t)sin^2(t) sound better?

7. See new attachment

again better double check

8. Originally Posted by Calculus26
Does 4 *sin^4(t) +16sin^2(t)sin^2(t) sound better?
I was thinking $\displaystyle \sqrt{ 4 \sin^4 \phi +16\sin^2 \phi \cos^2 \phi }$

9. What an idiot I am -- forgot the square root

10. Ok, I understand what you have done but have no Idea how to integrate $\displaystyle \sqrt{4sin^4(\theta)+16(cos(\theta)sin(\theta))^2}$
I tried reducing it to $\displaystyle sin(\theta)\sqrt{1+7cos^2\theta)}$
Then letting $\displaystyle u=cos(\theta)$ it becomes $\displaystyle \sqrt{1+7u^2}$ but I am really stuck at this point.

11. If my calculations are correct

4sin^4(t)+16cos^2(t)sin^2(t) = 4sin^2(t)[sin^2(t)+4cos^2(t)]

= 4sin^2(t)[1+3cos^2(t)]

taking sqrt roots

=2sin(t)[ 1 +3cos^2(t) ]^1/2

letting u = cos(t) you obtain

int(1+3u^2)^1/2du

Use the trig sub u = 1/[3^(1/2)]tan(a) 3^(1/2)du = sec^2(a)da

You get an integral of the form sec^3(a)