Substitutions in multiple integrals

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• Nov 22nd 2010, 09:14 PM
downthesun01
Substitutions in multiple integrals
R is a region in the first quadrant of the xy-plane bounded by xy=1, xy=9, y=x, and y=3x. Using the transformation $v=\sqrt{\frac{y}{x}}$, $u=\sqrt{xy}$ with u>0, v>0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G.

$\displaystyle \int\int_{R}(\sqrt{\frac{y}{x}}+\sqrt{xy})dxdy$

I'm totally lost on this. This is an exam review question but for all of our examples in the textbook the transformation is written in terms of x=... and y=...

I have no idea what to do here.
• Nov 22nd 2010, 09:54 PM
Chris L T521
Quote:

Originally Posted by downthesun01
R is a region in the first quadrant of the xy-plane bounded by xy=1, xy=9, y=x, and y=3x. Using the transformation $v=\sqrt{\frac{y}{x}}$, $u=\sqrt{xy}$ with u>0, v>0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G.

$\displaystyle \int\int_{R}(\sqrt{\frac{y}{x}}+\sqrt{xy})dxdy$

I'm totally lost on this. This is an exam review question but for all of our examples in the textbook the transformation is written in terms of x=... and y=...

I have no idea what to do here.

Observe that $xy=1\implies \sqrt{xy}=1$ and $xy=9\implies \sqrt{xy}=3$ (we can't allow for -3 given the restriction on u). Thus, we see that $1\leq u\leq 3$.

Similarly, we see that $y=x\implies \sqrt{\dfrac{y}{x}}=1$ and $y=3x\implies \sqrt{\dfrac{y}{x}}=\sqrt{3}$ (again, we can't allow for $-\sqrt{3}$ due to the restriction on v). Thus we see that $1\leq v\leq \sqrt{3}$. We now have the limits for the transformed integral.

Next, we need to get x and y in terms of u and v and then find its Jacobian.

Observe that

(1) $xv^2=y$ from the definition of v, and
(2) $u^2=xy$ from the definition of u.

Substitute (1) into (2) to see that $u^2=x^2v^2\implies x=\dfrac{u}{v}$.

Multiply the corresponding sides of (1) and (2) together to get $xu^2v^2=xy^2\implies y=uv$

Now find the Jacobian. I leave it for you to verify that $J(u,v)=\dfrac{2u}{v}$.

Therefore, $\displaystyle\iint\limits_R\left(\sqrt{\dfrac{y}{x }}+\sqrt{xy}\right)\,dx\,dy=\int_1^{\sqrt{3}}\int_ 1^3\left(2u+\frac{2u^2}{v}\right)\,du\,dv$

Can you proceed? Does this all make sense?
• Nov 22nd 2010, 10:18 PM
downthesun01
You helped tons. Thanks a bunch.