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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?

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  • $\begingroup$ Well, from the top of my head - it looks like you have $\pi$ placed wrong, at least. $\endgroup$
    – Kaster
    Nov 26, 2013 at 2:42

1 Answer 1

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$$ \frac 9{28} \int_0^{\frac \pi 2} \int_0^1 u \sin u^2 du dv = \frac {9 \pi}{112} \left[ 1 - \cos (1)\right] $$

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