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I hope that someone can help me determine the the bounds of integration for this problem.

Evaluate $$iintlimits_RxydA$$ where, $$R=(x,y): fracx^236+fracy^216leq1, xgeq0,ygeq0$$
my attempt, r=1, since $$(x,y): fracx^236+fracy^216leq1$$The region of integration is in the first quadrant since$$xgeq0,ygeq0$$I change from cartesion to polar $$f(x,y)=xy $$$$f(r,theta)=rcos(theta)rsin(theta)$$$$dA=rdrdtheta$$ So from everything above I thought the bounds would be r=0 to r=1 and $theta=0$ to $theta=fracpi2$ When I put everything together I get $$intlimits_0^fracpi2intlimits_0^1rcos(theta)rsin(theta)rdrdtheta$$
If I compute the integral above I get $frac18$, I know that $frac18$ is not the correct answer since the question is multiple choice and the options are (a) 5 (b) 25 (c) 55 (d) 72 (e) 73. I don't know where to go from here I assume that my error has to do with $(x,y): fracx^236+fracy^216leq1$ since I don't use $0leqyleq4$ and $0leqxleq6$ I'm just not sure how to incorporate the fractions are they part of r?

Any help would be greatly appreciated, thank you.

Let $x = 6rcostheta$ and $y =4rsintheta$

So, $fracx^26^2+fracy^24^2=r^2$

Now, $r^2le1$

$xge0$ and $yge0implies0le rle1$ and $0lethetalepi/2$

$dxdy = 24rdrdtheta$

$I = int^fracpi2_0int^1_0r^2(24sinthetacostheta)(24 r)dr dtheta = 576 int^fracpi2_0int^1_0r^2(sinthetacostheta)( r)dr dtheta = 576frac18 = 72$

(You've already found $frac18$, so I just multiplied it by $576$)

$x = 6rcostheta$ $y =4rsintheta$

Partial derivatives:

$x_r = 6costheta$, $y_r = 4sintheta$

$x_theta = - 6rsintheta$ , $y_theta = 4rcostheta $

$J = beginvmatrix6costheta & 4sintheta \ - 6rsintheta & 4rcosthetaendvmatrix = 24rcos^2theta+24rsin^2theta = 24r$

So, $dxdy = |J|drdtheta = 24r dr dtheta$

This question is a bit tricky because $R$ is not a circle quadrant, it's an ellipse quadrant. Therefore, the underlying ellipse needs to be transformed into a circle by a substitution – $u=frac23x$ works:
$$iint_Rxy,dA=frac94iint _Suy,dA$$
$S$ is now a quarter-disc of radius $4$ around the origin, so polar coordinates can now be used:
$$=frac94int_0^4int_0^pi/2r^3costhetasintheta,dtheta,dr$$
$$=frac94int_0^4frac12r^3,dr$$
$$=frac98×frac14×4^4=72$$

The entire problem is that the domain is 1 quarter of an ellipse, and not a circle. Thus, we must parameterize the domain as we would parameterize an ellipse.

As $fracx^236+fracy^216=1$ can be parameterized as $x=6costheta, y = 4sintheta$, the parameterization of the ellipse is $$x=6rcostheta, y = 4rsintheta$$ where $rle 1$, $theta in [0, fracpi 2]$

If you had $x^2+y^2=1$, then you would have a circle of radius $1$. If instead of $x^2+y^2=1$ you had $left(frac x rright)^2+left(frac y rright)^2=1$, then the radius would be $r$. But you for a circle, you can't have two different radii, so that leads to the conclusion that this can't be a circle in x-y coordinates; instead, it's an ellipse. To have a radius-1 circle, you need to do a change of coordinates: $u = frac x 6$ $v = frac y 4 $. Then you can do another change of coordinates to $r$ and $theta$ in terms of $u$ and $v$. Keep in mind that you need to take into account the Jacobians for both transformations. Or you can do it all in one transformation as in Ak19's answer.