Question

Evaluate the integral below, where C is the curver(t) = ‹sin(t),cos(t), sin(2t)›, 0 ≤ t ≤2π. (Hint: Observe that C lies on the surfacez = 2xy.)
∫C (y+9sin(x))dx +(z2+4cos(y))dy +x3dz

Answer 1

We can compute the integral directly: we have

$\begin{cases}x(t)=\sin t\\y(t)=\cos t\\z(t)=\sin(2t)\end{cases}\implies\begin{cases}\mathrm dx=\cos t\,\mathrm dt\\\mathrm dy=-\sin t\,\mathrm dt\\\mathrm dz=2\cos(2t)\,\mathrm dt\end{cases}$

Then the integral is

$\displaystyle\int_C(y+9\sin x)\,\mathrm dx+(z^2+4\cos y)\,\mathrm dy+x^3\,\mathrm dz$

$=\displaystyle\int_0^{2\pi}\bigg((\cos t+9\sin(\sin t))\cos t-(\sin^2(2t)+4\cos(\cos t))\sin t+\sin^3t\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\cos^2t+9\cos t\sin(\sin t)-\sin^2(2t)\sin t-4\sin t\cos(\cos t)+\sin^3t\,\mathrm dt$

You could also take advantage of Stokes' theorem, which says the line integral of a vector field $\vec F$ along a closed curve $C$ is equal to the surface integral of the curl of $\vec F$ over any surface $S$ that has $C$ as its boundary.

In this case, the underlying field is

$\vec F(x,y,z)=\langle y+9\sin x,z^2+4\cos y,x^3\rangle$

which has curl

$\mathrm{curl}\vec F(x,y,z)=-\langle2z,3x^2,1\rangle$

We can parameterize $S$ by

$\vec s(u,v)=\langle u\cos v,u\sin v,2u^2\cos v\sin v\rangle=\langle u\cos v,u\sin v,u^2\sin(2v)\rangle$

with $0\le u\le1$ and $0\le v\le2\pi$.

Note that when viewed from above, $C$ has negative orientation (a particle traveling on this path moves in a clockwise direction). Take the normal vector to $S$ to be pointing downward, given by

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle2u^2\sin v,2u^2\cos v,-u\rangle$

Then the integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S\mathrm{curl}\vec F(x,y,z)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1-\langle2u^2\sin(2v),3u^2\cos^2v,1\rangle\cdot\langle2u^2\sin v,2u^2\cos v,-u\rangle\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^14u^4\sin(2v)\sin v+6u^4\cos^3v-u\,\mathrm du\,\mathrm dv$

Both integrals are kind of tedious to compute, but personally I prefer the latter method. Either way, you end up with a value of $\boxed\pi$.