# Thread: Surface Area Question

1. ## Surface Area Question

Find the surface area of the cone z=4sqrt(x^2+y^2) and above a region in the xy-plane with area 3.
I tried this problem and here's what i've done------
r=sqrt(x^2+y^2)
The slope s is s=sqrt(r^2+16r^2)=sqrt(17)r

A surface=pir^2+pi r s=pi r^2(1+sqrt17)

3=
pi r^2(1+sqrt17) ===> r=0.4317

this is as far as i went, but what do I do next to get the surface area?

Thanks in advance!!!

2. ## Re: Surface Area Question

Have you learnt how to use cylindrical coordinates?
If so consider using
$\displaystyle x=rcos\theta$
$\displaystyle y=rsin\theta$
$\displaystyle z=z$

$\displaystyle dA= dxdy= |J|drd\theta$
where |J| is the determinant of the Jacobian

integrate dA between r=0 and r=0.4317 and between $\displaystyle \theta=0 and \theta=2\pi$

3. ## Re: Surface Area Question

Thanks! but I still don't get it is the determinant equals r? or 4r? I tried both and neither worked

4. ## Re: Surface Area Question

The jacobian is
$\displaystyle \begin{bmatrix}\frac{dx}{dr}&\frac{dx}{d\theta}\\ \frac{dy}{dr}&\frac{dy}{d\theta}\end{bmatrix}$

Which becomes
$\displaystyle \begin{bmatrix}cos\theta&-rsin\theta\\sin\theta&rcos\theta\end{bmatrix}$

The determinant is r, perhaps you are making a mistake elsewhere.