I'm a little stuck on a practice problem. In general, I know the steps, but I must be getting hung up somewhere, as I am not getting the same answer. Here is the problem:
$\iint_{D} xy dA$
Where D is the region bounded by the lines $y=x$, $y=x+4$, $x=0$ and $x=2$.Use the change of variables: $x=2u$ and y=2u$.
My first step is to draw the region. I won't put that here, but it ends up being a parallelogram. So I then use the transformations to find the Jacobian:
$\frac{dx}{du} \frac{dy}{dv} - \frac{dx}{dv} \frac{dy}{du} = 4$
Taking the partials for each I get (2*2) - 0, thus, 4. Then I need to figure out what my new equation will be using the transformations:
$xy \rightarrow (2u)(2u+4) = 4u^2+8uv $
So, I now have:
$\iint_{D*} (4u^2+8uv)4 du dv$
Now, I need to find my D* (new areas of integration). Using the transformation given, I translate the old areas into new:
$y=x \rightarrow 2u+v = 2u \rightarrow v=0$
$y=x+4 \rightarrow 2(2u+v) \rightarrow v=4$
$x=0 \rightarrow 2u=0 \rightarrow u=0$
$x=2 \rightarrow 2u=2 \rightarrow u=1$
So, this should result in limits of integration for v from 0 to 4 and for u from 0 to 1. Therefore, I should have the final integral to solve as begin:
$\int_0^4 \int_0^1 (4u^2+8uv)4 du dv$
However, when I solve this, I don't get the proper answer. I need to answer it in a decimal form, and that given answer is 26.6667. Where am I going wrong? Thanks for any help!
The suggested change of variables is incorrect. It should be $$x=2u,\quad y=2u+4v\qquad(0\leq u\leq1,\ 0\leq v\leq 1)\ .$$ Inspecting the figure one recognizes that for fixed $u\in[0,1]$ one obtains a constant $x\in[0,2]$, and when then $v$ moves from $0$ to $1$ the point $(x,y)$ moves vertically upwards in the drawn parallelogram, starting and ending and the desired $y$-values.
The Jacobian determinant then computes to $8$, and one has $\>x\,y=4u^2+8uv$. We therefore obtain $$\int_D x\,y\>{\rm d}(x,y)=\int_0^1\int_0^1 (4u^2+8uv)\>8\>dv\>du={80\over3}\ .$$