2 Attachment(s)

Transformations to evaluate an integral

Attachment 28785

The book prescribes the transformation T(*u,v*) = (*u*^{2} - *v*^{2}, 2*u**v*) where x = *u*^{2} - *v*^{2} and y = 2*u**v*. Using this information, I have found that the Jacobian of the transformation as 4(*u*^{2}* + v*^{2}), resulting in the following:

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I am just unable to find *R*, the region of integration. I graphed it in the *xy* plane looked like a the tip of a leaf protruding over the x-axis.

My biggest problem is that this *is not* a linear transformation and my professor hasn't taught us how to tackle those.

Any help will be greatly appreciated,

--Giest

Re: Transformations to evaluate an integral

You don't need to do a transformation at all...

If you draw the region and picture drawing horizontal strips, these strips are bounded on the left by $\displaystyle \displaystyle \begin{align*} \frac{y^2 - 4}{4} = x \end{align*}$ and bounded on the right by $\displaystyle \displaystyle \begin{align*} \frac{4 - y^2}{4} = x \end{align*}$, and then these strips are summed up between $\displaystyle \displaystyle \begin{align*} y = 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y = 2 \end{align*}$. So your integral is

$\displaystyle \displaystyle \begin{align*} \int_{0}^{2}{\int_{\frac{y^2 - 4}{4}}^{\frac{4-y^2}{4}}{y\,dx}\,dy} \end{align*}$

Re: Transformations to evaluate an integral

That thought certainly occurred to me, however, the course I need this for is a Linear Algebra course, so I am pretty sure my professor wants us to use linear transformations. I'll definitely use this to check my answer, though.

Re: Transformations to evaluate an integral

If you must use this transformation, when $\displaystyle \displaystyle \begin{align*} x = u^2 - v^2 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y = 2\,u\,v \end{align*}$, then that means your bounding functions become

$\displaystyle \displaystyle \begin{align*} y^2 &= 4 - 4x \\ \left( 2\, u\,v \right) ^2 &= 4 - 4 \left( u^2 - v^2 \right) \\ 4\, u^2 \, v^2 &= 4 - 4\, u^2 + 4 \, v^2 \\ u^2 \, v^2 &= 1 - u^2 + v^2 \\ u^2 + u^2 \, v^2 &= 1 + v^2 \\ u^2 \left( 1 + v^2 \right) &= 1 + v^2 \\ u^2 &= \frac{1 + v^2}{1 + v^2} \\ u^2 &= 1 \\ u &= \pm 1 \end{align*}$

and

$\displaystyle \displaystyle \begin{align*} y^2 &= 4 + 4x \\ \left( 2\, u \, v \right) ^2 &= 4 + 4 \left( u^2 - v^2 \right) \\ 4\,u^2\,v^2 &= 4 + 4\,u^2 - 4\,v^2 \\ u^2 \, v^2 &= 1 + u^2 - v^2 \\ v^2 + u^2\,v^2 &= 1 + u^2 \\ v^2 \left( 1 + u^2 \right) &= 1 + u^2 \\ v^2 &= \frac{1 + u^2}{1 + u^2} \\ v^2 &= 1 \\ v &= \pm 1 \end{align*}$

So your boundary is a square with u and v both bounded by $\displaystyle \displaystyle \begin{align*} \pm 1 \end{align*}$. Can you see what the bounds for your region must be now?

Re: Transformations to evaluate an integral

Ah, Thanks! Now that you have shown me, it's shamefully simple. I was just trying to map the vertices of the region from the xy plane to the uv plane, when, in stead, I should have been mapping the entire function.

Thanks, again!