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

`(\partial\vec s)/(\partial v)*(\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`.