Integrate x+y where the region is enclosed by y= 3 - x2 and y= 2x by polar parameterization (please only use Polar parameterization?

Sep 30, 2020
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Convert x+y to polar form
Y = 3 - X ^2 => X^2 = 3 - Y = -(Y - 3)
X^2 = -(Y- 3) =>
r2 cos 2 (theta) + r sin(theta) - 3 = 0
r2 sin 2 (theta) - r sin (theta) + 3 - r2 = 0
Sin ( theta ) = ( r +/- sqrt( r2 - 4*r2*(3-r2)) )/ 2r2
Sin (theta) = ( 1 +/ - sqrt ( 4r2 - 11) ) / 2r
That is equation of parabola whose center is
(0, 3)
And focus lies on negative Y direction, therefore equation of line encloses curve in Ist quadrant, 2nd quadrant, 3rd quadrant. For
Now in polar form given line
Y = 2X => y / x = 2 => tan (theta) = 2
theta = tan -1 (2)
Therefore sum transforms in polar form
Theta = tan -1 (2) to
Theta = sin -1 { (1 +/- sqrt ( 4r2 - 11 ) ) / 2r }
r ( cos theta + sin theta )
Now calculate simple integration wrt theta.
 

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