Let $\displaystyle \Phi(u,v) = (u-v,u+v,u)$ , and let D be the unit disk in the uv plane. Find the area of $\displaystyle \Phi(D)$
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.
Let $\displaystyle \Phi(u,v) = (u-v,u+v,u)$ , and let D be the unit disk in the uv plane. Find the area of $\displaystyle \Phi(D)$
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.
You said "I've parametrized u and v" so I assume you have $\displaystyle u= r cos(\theta)$, $\displaystyle v= r sin(\theta)$. Put those in for u and v and integrate with r going from 0 to 1 and $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$ 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.)
It's a linear transformation
$\displaystyle T(\begin{bmatrix}u\\v \end{bmatrix})=\begin{bmatrix}u-v\\u+v\\u \end{bmatrix}\rightarrow T(\vec{e}_1)=\begin{bmatrix}1\\1\\1 \end{bmatrix}~,~
T(\vec{e}_2)=\begin{bmatrix}-1\\1\\0 \end{bmatrix}$
$\displaystyle 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 $