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In this problem, we are asked to evaluate the double integral over the region r x squared da, where the region r is the ellipse given by 9x squared plus 4y squared equals to 36.
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We are asked to evaluate the double integral by making the transformation x equals to 2 times u and y equals to 3 times v.
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So here, let us substitute the value of xs to u and y as 3v.
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We obtain 9 times 4 u squared plus 4 times 9 v squared equals to 36.
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Dividing both sides by 36 we get u squared plus v squared equals to 1 and this is a unit circle.
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For a unit circle in polar coordinates we have r to be varying from 0 to 2.
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1 and we have tita to vary from 0 to 2 times pi.
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So further we require the jacobian.
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So the jacobian is given by the determinant of x u, xv, y u, yv.
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So we have the determinant of x partially differentiated with respect to u is 2, partially differentiated with respect to v is 0, y partially differentiated with respect to u is 0, and y partially differentiated with respect to v is 3.
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So we have the determinant to be equal to 6 minus 0, which implies that the jacobian is 6.
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So now let us find out the double integral.
01:46
The double integral over the region are x squared d a equal to the double integral over the region are substituting x as 2 u we have 4 u squared times a jacobian which is 6 times d v d u...