Find $ \int_C (2xy^3+\cos x)\,dx + (3x^2y^2+5x)\,dy $ where $C$ is $x^2+y^2=64$

Question

I want to evaluate the line integral $$ \int_C (2xy^3+\cos x)\,dx + (3x^2y^2+5x)\,dy $$ where $C$ is the circle $x^2+y^2=64$

I parametrized the circle by $r(t)=8\cos t \ \hat{i} +8\sin t \ \hat{j}$ for $0≤t≤2\pi$ but not sure what to do next

Answer 1

which stops you? I suppose if you notice $$ \int_{C} \cos x dx=0, $$ the rest would be easier.

Answer 2

Hint: $2xy^3 \ dx$ and $3x^2y^2 \ dy$ look suspiciously similar; in fact they can be written as parts of an exact differential of $x^2y^3$. Likewise for $\cos x \ dx = \cos x \ dx + 0 \ dy$.

In other words, if $f(x,y) = x^2 y^3 + \sin x \ $ then $df = (2xy^3 + \cos x)\ dx + 3x^2y^3 \ dy \ $ and hence

$$\int_C (2xy^3 + \cos x)\ dx + 3x^2y^3 \ dy = 0$$

for any closed path $C$.

So your integral simplifies significantly

$$\int_C (2xy^3 + \cos x)\ dx + (3x^2y^3 + 5x) \ dy = \int_C 5x \ dy$$

To evaluate $\displaystyle \int_C 5x \ dy \ $, you can use the substitutions you originally proposed. Note that the orientation of the circle matters, so double check the original question for whether or not you are supposed to go clockwise or anticlockwise.

Answer 3

You can also use Green theorem, which says

$$\int_{\partial D}Pdx+Qdy =\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy.$$

Thus \begin{align} \int_C (2xy^3+\cos x)dx + (3x^2y^2+5x)dy &= \int_B (6xy^2+5-6xy^2)dxdy\\ &=\int_0^8\int_0^{2\pi} 5r dr d\theta\\ &=\frac{5\cdot 8^2}{2}\cdot 2\pi\\ &=320\pi, \end{align} where $B=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le 8^2\}$.