How Can I Calculate a Double Integral Using a Change of Variable?

[mathmari]

Hey!

Defining $u=x^2+y^2$, $v=x^2+y^2-2y$ I want to calculate $\iint_D xe^y dxdy$ using the change of variable, where $D$is the space that is determined by the inequalities $x\geq 0$, $y\geq 0$, $x^2+y^2\leq 1$ and $x^2+y^2\geq 2y$. I have done the following:

We have the new variables $u=x^2+y^2$, $v=x^2+y^2-2y$.

When we solve for $x,y$ as a function of $u,v$ we get:
$$v=x^2+y^2-2y\Rightarrow v^2=u^2-2y \Rightarrow y=\frac{u}{2}-\frac{v}{2}$$
$$u=x^2+y^2 \Rightarrow x^2=u-y^2 \Rightarrow x^2=u-\left (\frac{u-v}{2}\right )^2 \Rightarrow x^2=u-\frac{u^2-2uv+v^2}{4}
\\ \Rightarrow x^2=\frac{4u-u^2+2uv-v^2}{4} \Rightarrow x=\pm \sqrt{\frac{4u-u^2+2uv-v^2}{4}} \\ \Rightarrow x=\pm \frac{\sqrt{4u-u^2+2uv-v^2}}{2}$$ Since $x\geq 0$ we get $x= \frac{\sqrt{4u-u^2+2uv-v^2}}{2}$.

Is everything correct so far? (Wondering) Now we have to calculate the determinant, right? We get
\begin{align*}det \begin{pmatrix}\frac{\partial{x}}{\partial{u}} & \frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}} & \frac{\partial{y}}{\partial{v}}\end{pmatrix}&=det \begin{pmatrix}\frac{2-u+v}{2\sqrt{4u-u^2+2uv-v^2}} & \frac{u-v}{2\sqrt{4u-u^2+2uv-v^2}} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix} \\ & =-\frac{2-u+v}{4\sqrt{4u-u^2+2uv-v^2}}-\frac{u-v}{4\sqrt{4u-u^2+2uv-v^2}} \\ & =-\frac{2-u+v+u-v}{4\sqrt{4u-u^2+2uv-v^2}} \\ & =-\frac{1}{2\sqrt{4u-u^2+2uv-v^2}}\end{align*}

Therefore we get the following:
\begin{align*}\iint_D xe^y dxdy&=\iint_E \frac{\sqrt{4u-u^2+2uv-v^2}}{2}e^{\frac{u-v}{2}} \left |\frac{\partial{(x,y)}}{\partial{(u,v)}}\right |dudv \\ & =\iint_E \frac{\sqrt{4u-u^2+2uv-v^2}}{2}e^{\frac{u-v}{2}}\frac{1}{2\sqrt{4u-u^2+2uv-v^2}}dudv \\ & =\frac{1}{4}\iint_E e^{\frac{u-v}{2}}dudv\end{align*}

Is the new space $E=\{(u,v) \mid 0\leq u\leq 1, 0\leq v\leq 1\}$ ? I thought so because of the following:
Since $x^2+y^2\leq 1 \Rightarrow u\leq 1$. Since $x\geq 0, y\geq 0$ we have that $x^2+y^2\geq 0$ and so $u\geq 0$.
Since $x^2+y^2\geq 2y \Rightarrow v\geq 0$. Since $x^2+y^2\leq 1$ and $y\geq 0$ we have that $x^2+y^2-2y\leq 1-2\cdot 0=1$ and so $0\leq v\leq 1$.

(Wondering)

 

[I like Serena]

We have the new variables $u=x^2+y^2$, $v=x^2+y^2-2y$.

When we solve for $x,y$ as a function of $u,v$ we get:
$$v=x^2+y^2-2y\Rightarrow v^2=u^2-2y \Rightarrow y=\frac{u}{2}-\frac{v}{2}$$

Hey mathmari! (Smile)

Shouldn't there be a couple of squares in there? (Wondering)

 

[mathmari]

Shouldn't there be a couple of squares in there? (Wondering)

Oh (Blush) Shouldn't it be as follows?
$$v=x^2+y^2-2y\Rightarrow v=u-2y \Rightarrow y=\frac{u}{2}-\frac{v}{2}$$
(Wondering)

 

[I like Serena]

Oh (Blush) Shouldn't it be as follows?
$$v=x^2+y^2-2y\Rightarrow v=u-2y \Rightarrow y=\frac{u}{2}-\frac{v}{2}$$

Ah yes, with that correction it looks correct to me.

Now we have to calculate the determinant, right?

Yep. I didn't check the calculations, but overall it looks correct.

Is the new space $E=\{(u,v) \mid 0\leq u\leq 1, 0\leq v\leq 1\}$ ? I thought so because of the following:
Since $x^2+y^2\leq 1 \Rightarrow u\leq 1$. Since $x\geq 0, y\geq 0$ we have that $x^2+y^2\geq 0$ and so $u\geq 0$.
Since $x^2+y^2\geq 2y \Rightarrow v\geq 0$. Since $x^2+y^2\leq 1$ and $y\geq 0$ we have that $x^2+y^2-2y\leq 1-2\cdot 0=1$ and so $0\leq v\leq 1$.

Can we draw $D$?
That should tells us if it is likely that it gets mapped to a square. (Wondering)

 

[mathmari]

Can we draw $D$?
That should tells us if it is likely that it gets mapped to a square. (Wondering)

$D$ is the following:

[DESMOS=-2,2,-2,2]x^2+y^2\ge2y\ \left\{x^2+y^2\le1\ \left\{x\ge0\right\}\ \left\{y\ge0\right\}\right\}[/DESMOS]So will $E$ be a triangle? (Wondering)

We have the following:
$$2y\leq x^2+y^2\leq 1 \Rightarrow 2y\leq 1\Rightarrow y\leq \frac{1}{2}$$
Therefore we get $$0\leq y\leq \frac{1}{2} \Rightarrow 0\leq \frac{u-v}{2}\leq \frac{1}{2} \Rightarrow 0\leq u-v\leq 1 \Rightarrow -u\leq -v\leq 1-u \Rightarrow u-1\leq v\leq u$$

And from $0\leq x^2+y^2\leq 1$ we get $0\leq u\leq 1$.

Using that bounds I got $\frac{e^{\frac{1}{2}}-1}{2}$.

Is this correct? (Wondering)

 

[I like Serena]

$D$ is the following:

So will $E$ be a triangle? (Wondering)

Looks like it yes.

We have the following:
$$2y\leq x^2+y^2\leq 1 \Rightarrow 2y\leq 1\Rightarrow y\leq \frac{1}{2}$$
Therefore we get $$0\leq y\leq \frac{1}{2} \Rightarrow 0\leq \frac{u-v}{2}\leq \frac{1}{2} \Rightarrow 0\leq u-v\leq 1 \Rightarrow -u\leq -v\leq 1-u \Rightarrow u-1\leq v\leq u$$

And from $0\leq x^2+y^2\leq 1$ we get $0\leq u\leq 1$.

Using that bounds I got $\frac{e^{\frac{1}{2}}-1}{2}$.

Is this correct? (Wondering)

Which bounds for $v$ exactly?
$u-1\leq v\leq u$ doesn't look correct to me.
After all, that would mean that for $u=0$ we have $-1\le v\le 0$, but we should have $v\ge 0$.

Generally, we should check where each of the corners map to. That is, (0,0), (1,0), and $(\frac 12\sqrt 3, \frac 12)$.
And we should check what each bounding curve maps to. That is, the x-axis, the circle with radius 1 and center (0,0), and the circle with radius 1 and center (0,1).
Drawing the results should show us what we have exactly. (Thinking)