Question

Evaluate double integral `int int (1-x^2/a^2-y^2/b^2)dx dy` over the first quadrant of the ellipse `x^2/a^2+y^2/b^2 = 1`, by using the transformation `x=au` and `y=bv`?

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Answer 1

`(piab)/8`

Explanation:

We seek:

`I = int \ int_R \ (1-x^2/a^2-y^2/b^2) \ dx \ dy`

Where:

`R={(x,y) | x^2/a^2+y^2/b^2 le 1, x,y ge 0}`

Applying the suggested transformation:

`x = au`
`y = bv`

Then we can transform the region of integration:

`S = {(u,v) | (au)^2/a^2 + (bv)^2/b^2 le 1, u,v ge 0}`
`\ \ = {(u,v) | u^2 + v^2 le 1, u,v ge 0}`

And we similarly transform the double integral, and apply the Jacobian:

`I = int \ int_S \ (1-(au)^2/a^2-(bv)^2/b^2) \ | \ du \ dv`

`\ \ = int \ int_S \ (1-u^2-v^2) \ |(partial(x,y))/(partial(u,v)) | \ du \ dv`

And we compute the Jacobian:

`J = |(partial(x,y))/(partial(u,v)) | = | ( (partial x)/(partial u), (partial x)/(partial v)) , ((partial y)/(partial u),(partial y)/(partial v)) | = | ( a, 0) , (0,b) | = ab`

So then:

`I = int \ int_S \ (1-u^2-v^2) \ ab \ du \ dv`

We are now integrating over the first quadrant of a unit circle, and so a further transformation to polar coordinates is appropriate:

`u = rcos theta`
`v = r sin theta`

So we have:

`C = {(r,theta) | 0 le r le 1, 0 le theta le pi/2}`

`I = int \ int_C \ (1-(rcos theta)^2-(rsin theta)^2) \ ab \ r \ dr \ d theta`

`\ \ = int \ int_C \ (1-r^2) \ ab \ r \ dr \ d theta`

`\ \ = int_0^(pi/2) \ int_0^1 \ (1-r^2) \ ab \ r \ dr \ d theta`

If we consider the inner integral:

`int_0^1 \ (1-r^2) \ ab \ r \ dr = ab \ int_0^1 \ r-r^3 \ dr`

`" " = ab \ [r^2/2-r^4/4]_0^1`

`" " = ab \ (1/2-1/4)`

`" " = (ab)/4`

So that:

`I = int_0^(pi/2) (ab)/4 d theta`

`\ \ = [(ab)/4 theta]_0^(pi/2)`

`\ \ = (ab)/4pi/2`

`\ \ = (piab)/8`