# Math Help - Transformations to evaluate an integral

1. ## Transformations to evaluate an integral

The book prescribes the transformation T(u,v) = (u2 - v2, 2uv) where x = u2 - v2 and y = 2uv. Using this information, I have found that the Jacobian of the transformation as 4(u2 + v2), resulting in the following:

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

2. ## 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 \begin{align*} \frac{y^2 - 4}{4} = x \end{align*} and bounded on the right by \displaystyle \begin{align*} \frac{4 - y^2}{4} = x \end{align*}, and then these strips are summed up between \displaystyle \begin{align*} y = 0 \end{align*} and \displaystyle \begin{align*} y = 2 \end{align*}. So your integral is

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

3. ## 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.

4. ## Re: Transformations to evaluate an integral

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

\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 \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 \begin{align*} \pm 1 \end{align*}. Can you see what the bounds for your region must be now?

5. ## 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!