Area of the region which is the interior of the curve r9cost

Area of the region which is the interior of the curve r=9cos(t) as t varies from t= -pi/2 to t= pi/2

Since cos(t) is symmetric about y axis we need to integrate the function from 0 to pi/2 and multiply the answer by 2 since integral of cos(t) is sin(t) area bounded will be = 9*2*(sin(pi/2)-sin(0)) =18 units