Question

Evaluate the double integral by polar coordinates. Integral D 8- 2x² - 2y²dA,

Answer 1

$\iint(8-2x^2-2y^2)dA\\ =\iint(8-2x^2-2y^2)(r)(d\theta)(dr)\\ =\iint(8-2(x^2+y^2))(r)(d\theta)(dr)$

Put $x=rcos\theta,y=rsin\theta$

$\therefore x^2+y^2=r^2$

So, double integral us reduced to:

$\iint(8-2r^2)(r)(d\theta)(dr)\\ =\iint(8r-2r^3)(d\theta)(dr)\\ =\int(4r^2-\frac{r^4}{2}+c)(d\theta)\\ =4r^3\theta-\frac{r^4\theta }{2}+c\theta+d \\$

where $c,d$ are constants of integration.