Question

Evaluate integral of c (xy^4) ds where c is right half of the circle x^2 + y^2 =16 traced out in a counter clockwise direction parameterized using x=4cost, y=4sint

Answer 1

We can parametrize $C$ by

$$\vec r=\langle4\cos t, 4\sin t\rangle, -\pi/2\le t\le \pi/2$$

Then we have

$$\vec r'=\langle-4\sin t, 4\cos t\rangle$$

$$|\vec r'|=\sqrt{(-4\sin t)^2+(4\cos t)^2}=4$$

$$\int_Cxy^4ds=\displaystyle\int_{-\pi/2}^{\pi/2}(4\cos t)(4\sin t)^4(4)dt$$

$$=4^6[\dfrac{\sin ^5t}{5}]\begin{matrix} \pi/2 \\ -\pi/2 \end{matrix}=\dfrac{2^{13}}{5}$$