Define $T(u,v)=(u^2-v^2,2uv)$. Let $D^*$ be the set of $(u,v)$ with $u^2+v^2\leq1,u\geq0,v\geq0$. Find $T(D^*)=D$. Evaluate $\iint_D dxdy$.
We know $D^*=\{(\cos t,\sin t): 0\leq t\leq\pi/2\}$ the fourth part of the unit circle in the first quadrant. Therefore $T(D^*)=T(\cos t, \sin t) =(\cos 2t, \sin 2t)$ extends $D^*$ to cover the whole upper part of the unit circle.
Since $\textbf{D}T= \begin{pmatrix} 2u & -2v \\ 2v & 2u\end{pmatrix}$, the Jacobian determinant is $4u^2+4v^2$ which is only zero at a boundary point of the region of integration (a measure zero set; irrelevant for integration).
Hence, we define
$$\int_0^1 \int_0^{\sqrt{1-v^2}} 4u^2 + 4v^2 dudv \stackrel{\text{substitutions...}}{=} \frac{4}{3} \int_0^1 (1-v^2)^{3/2}dv$$
Now I don't know how to find this integral.
Do I have to map $D^*$ first to polar coordinates and then find its image under $T$? This would mean computing two Jacobians, right?
I'm really lost.
To do the integral, make the substitution $v=\sin\phi.$ Then it takes the form $$\int_0^{π/2}{\cos^4\phi\mathrm d \phi}.$$ To do this, note that $\cos^4\phi-\sin^4\phi=\cos 2\phi$ and $$\cos^4\phi+\sin^4\phi=\frac34+\frac14\cos 4\phi.$$