: Double integral(3sin(9x^2 81y^2))dA where R is the region in the first quadrant bounded by the ellipse 9x^2 81y^2=1

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: Double integral(3sin(9x^2 81y^2))dA where R is the region in the first quadrant bounded by the ellipse 9x^2 81y^2=1

Mathematics
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there is plus sign in between 9x^2 and 81y^2
Try the change of variables x = 1/3rcos(w), y = 1/2rsin(w). In terms of the (r,w) variables the region R is just r = 1, w = 0 to pi/2 (first quadrant) since 9x^2 + 4y^2 = r^2. The double integral in terms of (r,w) becomes int_{w=0}^{w=pi/2}_{r=0}^{r=1} sin(r^2) J([x,y],[r,w]) dr dw where J([x,y],[r,w]) is the determinant of the 2 x 2 Jacobian matrix partial dx/partial dr partial dx/partial dw partial dy/partial dr partial dy/partial dw which here is 1/3cos(w) -1/3rsin(w) 1/2sin(w) 1/2rcos(w) so the determinant J([x,y], [r,w]) is 1/6r(cos^2(w) + sin^2(w)) = 1/6r. The integral becomes 1/6 int_{w=0}^{w=pi/2}_{r=0}^{r=1} sin(r^2) r dr dw = 1/6 [w]_{w=0}^{w=pi/2} [-1/2cos(r^2)]_{r=0}^{r=1} = -1/12[pi/2-0][cos(1) - cos(0)] = 1/12(pi/2)(1 - cos(1) = pi/24*(1 - cos(1)).
how did you type that so fast? haah but thanks

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so is pi/24(1-cos(1)) the final answer?

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