# evaluating line integral (multivariable calculus)

Evaluate the line integral $\int_C y(x^2+y^2)dx-x(x^2+y^2)dy+xydz$ where $C$ is parametrized by $r(t)=\cos t i+\sin t j+tk$ for $-\pi\leq t\leq \pi$.

If I did it right (which I'm not sure if I did), then I would have gotten integral from $-\pi$ to $\pi$ of $(-1+\sin t \cos t)$...but how would I integrate the $\sin t \cos t$ part? Thanks.

Edit: I got -2$\pi$ -1 as an answer...is that correct?

## 1 Answer

we have that $$\int_Cy(x^2+y^2)dx-x(x^2+y^2)dy+xydz\\ x=\cos t,dx=-\sin tdt\\ y=\sin t,dy=\cos tdt\\ z=t,dz=dt\\ =\int_{-\pi}^{+\pi}-\sin^2t(\cos^2t+\sin^2t)-\cos^2t(\cos^2t+\sin^2t)+\cos t\sin tdt\\ =\int_{-\pi}^{+\pi}-\sin^2t-\cos^2t+\cos t\sin tdt\\ =\int_{-\pi}^{+\pi}\cos t\sin t-1dt$$ you can split into two integral $$=\int_{-\pi}^{+\pi}\cos t\sin tdt-\int_{-\pi}^{+\pi}dt$$ for the first you can make $u=\sin t,du=\cos tdt$. $$=\int_{0}^{0}udu-t\bigg|_{-\pi}^{+\pi}\\ =\frac{1}{2}u^2\bigg|_{0}^{0}-(\pi--\pi)\\ =\frac{1}{2}\sin^2t\bigg|_{-\pi}^{+\pi}-(\pi+\pi)\\ =-2\pi$$