Is This Week's Multivariable Calculus Challenge Doable for You?

[Chris L T521]

Thanks to those who participated in last week's POTW. I'll just say it's no fun unless more people participate!

I got some complaints about the difficulty of some of the previous problems; so, for this week's problem, I've decided to go with something a "tad" easier and doable -- a problem from multivariable calculus.

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Problem: Evaluate

\[\iint\limits_R \sin(xy)\,dA\]by making an appropriate change of variables where $R$ is the region enclosed by the curves $xy=\pi$, $xy=2\pi$, $xy^4=1$ and $xy^4=2$.

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I will provide a hint/suggestion for this week's problem:

Spoiler

Use the fact that if $\displaystyle\frac{\partial(x,y)}{\partial(u,v)}= \begin{vmatrix} \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}$ is the Jacobian of our transformation, then $\dfrac{\partial(x,y)}{\partial(u,v)} = \dfrac{1}{\partial(u,v)/\partial(x,y)}$.

This form of the Jacobian comes in handy when we can't explicitly solve for $x$ and $y$ in terms of $u$ and $v$.

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-(POTW)-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

The week six POTW was correctly answered by Sudharaka. Here's my solution:

Spoiler

Let $u=xy$ and $v=xy^4$. Then it follows that $\pi\leq u\leq 2\pi$ and $1\leq v\leq 2$. In this situation, it's not possible to explicitly solve for $x$ and $y$ in terms of $u$ and $v$. This requires us to apply the alternate definition for the Jacobian of our transformation. We see that

\[\frac{\partial(u,v)}{\partial(x,y)}= \begin{vmatrix} y & y^4 \\ x & 4xy^3\end{vmatrix} = 3xy^4 = 3v\]

and thus

\[\frac{\partial(x,y)}{\partial(u,v)}= \frac{1}{ \partial(u,v) /\partial(x,y) } = \frac{1}{3v}\]

Therefore,

\[\iint\limits_R \sin(xy)\,dA = \int_1^2\int_{\pi}^{2\pi}\frac{\sin u}{3v}\,du\,dv= \left(\int_1^2\frac{1}{3v}\,dv\right) \left(\int_{\pi}^{2\pi}\sin u\,du\right) =-\frac{2}{3}\ln(2).\]