I have a homework problem, as follows:
Evaluate the double integral by making an appropriate change of variables.
$\iint_R 9\sin(49x^2+16y^2)\,dA$, where $R$ is the region in the first quadrant bounded by the ellipse $49x^2 + 16y^2 = 1$
My work is as follows:
- Let $x\overset{\Delta}=\tfrac17u\cos v$ and $y\overset\Delta=\tfrac14u\sin v$.
- The Jacobian is the determinant $$|J|=\begin{Vmatrix} \tfrac17\cos v & -\tfrac17u\sin v\\ \tfrac14\sin v & \tfrac14u\cos v \end{Vmatrix}=28u\cos^2v+28u\sin^2v=\tfrac u{28}\\ $$
- The limits of integration are $u\in[0,1]$, $v\in[0,\pi/2]$.
- The integral is $$\frac9{28}\int_0^{\pi/2}\int_0^1u\sin u^2\,du\,dv$$ which evaluates to $\frac9{112\pi}(1-\cos(1))$.
However, my WebAssign (online grading system) says that this is incorrect.
Where have I gone wrong?
I would be happy to provide more detailed work for any step if it would be helpful or necessary.
EDIT: I've done some additional research and found a couple questions on the world's most authoritative source for math help that seem to give similar answers (here and here); am I missing something?
$$ \frac 9{28} \int_0^{\frac \pi 2} \int_0^1 u \sin u^2 du dv = \frac {9 \pi}{112} \left[ 1 - \cos (1)\right] $$