Answer
$125 \pi$
Work Step by Step
In order to compute the integral we will use the following formula:
$\oint F \ dr=\int_m^{n} F[r(t)] r'(t) \ dt$
Here, the curve is given as: $r(t)= (5 \cos t, 5 \sin t)$
This implies that $r'(t) = (-5 \sin t, 5 \cos t) \implies |r'(t)|=\sqrt {(-5 \sin t)^2+(5 \cos t)^2}=5$
Therefore, the integral is:
$\oint F \ dr=\int_0^{\pi} F[r(t)] r'(t) \ dt\\=\int_0^{\pi} (25 \sin^2 t-(2) (5\sin t)(5 \cos t)+25 \cos^2 t) \cdot 5 \ dt\\=5 \times \int_0^{ \pi} (25-50 \sin t \cos t) \ dt \\=5 \times [25(t)+25 \cos^2 t]_0^{\pi}\\=125 \pi $
