Question

Evaluate the line integral

6 integral yzcosx ds x = t, y = cost, z = sint 0 < t < pi

Answer 1

It looks like the integral is

`\displaystyle 6 \int yz\cos(x) \, ds`

With
`x=t`,
`y=\cos(t)`, and
`z=\sin(t)`, the line element is

`ds = \sqrt{\left((dx)/(dt)\right)^2 + \left((dy)/(dt)\right)^2 + \left((dz)/(dt)\right)^2} \, dt = \sqrt2 \, dt`

and so

`\displaystyle 6 \int yz\cos(x) \, ds = 6\sqrt2 \int_0^\pi \cos^2(t) \sin(t) \, dt`

Substitute
`u = \cos(t)` and
`du=-\sin(t)\,dt`, then

`\displaystyle 6 \int yz\cos(x) \, ds = -6\sqrt2 \int_1^(-1) u^2 \, du = 12\sqrt2 \int_0^1 u^2 \, du = \boxed{4\sqrt2}`

where in the second-to-last equality, we have

`\displaystyle -\int_1^(-1) = \int_(-1)^1`

and

`\displaystyle \int_(-1)^1 u^2 \, du = 2 \int_0^1 u^2\,du`

by symmetry.