Evaluating double integrals using the change of variable formula

Question

Evaluate

$$\iint_{D} \frac{dxdy}{\sqrt{1+x+2y}}$$ where $D=[0,1]\times[0,1]$, by setting $T(u,v)=(u,v/2)$ and evaluating and integral over $D^*$ such that $T(D^*)=D$.

I'm having a hard time with this one!

Since $T$ is linear, $T^{-1}$ is defined,

$$T^{-1}= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$$

and $D^*=T^{-1}(D)=[0,1] \times [0,2]$. We're mapping the $uv$-plane to the $xy$-plane, so the Jacobian should have the derivatives of $x,y$ with respect to $u,v$:

$$JT= \begin{vmatrix} 1 & 0 \\ 0 & 1/2 \\ \end{vmatrix} = 1/2$$

Hence, our integral is,

$$\frac{1}{2}\int^{2}_{0}\int^{1}_{0} \frac{dudv}{\sqrt{1+u+v}}$$

Is this correct? How can I find a primitive for this integrand?

Answer 1

HINT

Note that $$ \int \frac{da}{\sqrt{b+a}} = \int (b+a)^{-1/2} da = \frac{(b+a)^{1/2}}{1/2} +C, $$ where $b\geq 0$.

Answer 2

Yes your transformation is right. To do the computation, recall that you first integrate with respect to $u,$ keeping $v$ fixed. Thus, if you set $a=1+v$ the first integral is $$\int_0^1\frac{\mathrm d u}{\sqrt{a+u}}.$$ Can you continue now? (Hint: Use the transformation $a+u=z,$ and recall that you may write the integrand as $(a+u)^{-1/2}.$)