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anonymous
 one year ago
: 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
anonymous
 one year ago
: 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is plus sign in between 9x^2 and 81y^2

Jacob902
 one year ago
Best ResponseYou've already chosen the best response.0Try 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/20][cos(1)  cos(0)] = 1/12(pi/2)(1  cos(1) = pi/24*(1  cos(1)).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you type that so fast? haah but thanks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is pi/24(1cos(1)) the final answer?
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