# Thread: Area of a Surface

1. ## Area of a Surface

Let $\Phi(u,v) = (u-v,u+v,u)$ , and let D be the unit disk in the uv plane. Find the area of $\Phi(D)$

I've parametrized u and v but i dont see how to put the angles in since it's asking for the area of the sphere.

2. Originally Posted by larryboi7
Let $\Phi(u,v) = (u-v,u+v,u)$ , and let D be the unit disk in the uv plane. Find the area of $\Phi(D)$

I've parametrized u and v but i dont see how to put the angles in since it's asking for the area of the sphere.
The Surface area is $\displaystyle \iint_{D}\bigg| \bigg| \frac{\partial \Phi}{\partial u} \times \frac{\partial \Phi}{\partial v}\bigg| \bigg|dudv$

3. Originally Posted by TheEmptySet
The Surface area is $\displaystyle \iint_{D}\bigg| \bigg| \frac{\partial \Phi}{\partial u} \times \frac{\partial \Phi}{\partial v}\bigg| \bigg|dudv$
Thank you but i already know the equation it's the setting it up part.

4. You said "I've parametrized u and v" so I assume you have $u= r cos(\theta)$, $v= r sin(\theta)$. Put those in for u and v and integrate with r going from 0 to 1 and $\theta$ from 0 to $2\pi$ in order to cover the unit circle.

(No, it's not "asking for the area of the sphere". It is asking for the area of a region on the given plane.)

5. It's a linear transformation

$T(\begin{bmatrix}u\\v \end{bmatrix})=\begin{bmatrix}u-v\\u+v\\u \end{bmatrix}\rightarrow T(\vec{e}_1)=\begin{bmatrix}1\\1\\1 \end{bmatrix}~,~
T(\vec{e}_2)=\begin{bmatrix}-1\\1\\0 \end{bmatrix}$

$A=\begin{bmatrix}1&-1\\1&1\\1&0 \end{bmatrix}~,~S_{\Phi}=\sqrt{\det (A^TA)}\cdot S_{D}=\sqrt{6}~\pi r^2=\sqrt{6}~\pi$

6. Thank You much appreciation.