For those of us with monitors on which the above causes horizontal scroll, the image displays something along these lines:
Evaluate the line integral:
\(\displaystyle \L \int_C\, y\, dx\, +\, z\, dy\, +\, x\, dz\, \mbox{ for }\, t\, \in \, \[0,\, 2\pi\]\)
--------------------------------------------------
\(\displaystyle \L x\, =\, \sin{(t)}\, \mbox{ so } \, dx\, =\, \cos{(t)}\, dt\)
\(\displaystyle \L y\, =\, 2\sin{(t)}\, \mbox{ so }\, dy\, =\, 2\cos{(t)}\, dt\)
\(\displaystyle \L z\, =\, \sin^2{(t)}\, \mbox{ so }\, dz\, =\, 2\sin{(t)}\cos{(t)}\, dt\)
--------------------------------------------------
\(\displaystyle \L \int_0^{2\pi}\, \left[\, 2\sin{(t)}\cos{(t)}\, dt\, +\, \sin^2{(t)}2\cos{(t)}\, dt\, +\, \sin{(t)}2\sin{(t)}\cos{(t)}\, dt\, \right]\)
\(\displaystyle \L \int_0^{2\pi}\, \left[\, 2\sin{(t)}\cos{(t)}\, +\, 2\sin^2{(t)}\cos{(t)}\, +\, 2\sin^2{(t)}\cos{(t)}\, \right]\, dt\)
--------------------------------------------------
\(\displaystyle \L \underbrace{\int_0^{2\pi}\, \left[ \, 2\sin{(t)}\cos{(t)}\, \right]}_{\mbox{I}} \, dt\, +\, \underbrace{\int_0^{2\pi}\, \left[\, 2\sin^2{(t)}\cos{(t)}\, \right]}_{\mbox{II}}\, dt\, +\, \underbrace{\int_0^{2\pi}\, \left[\, 2\sin^2{(t)}\cos{(t)}\, \right]\, dt}_{\mbox{III}}\)
--------------------------------------------------
integral I:
\(\displaystyle \L \int_0^{2\pi}\, \left[ \, 2\sin{(t)}\cos{(t)}\, \right]\, =\, 2\, \int_0^{2\pi}\, \sin{(t)}\cos{(t)}\, dt\,\)
\(\displaystyle \L =\, 2\int_a^b\, u\, du\, =\, 2\, \left[\frac{u^2}{2}\right]\, =\, u^2\, |_a^b\, =\, \sin^2{(t)}\,|_0^{2\pi}\, =\, 0\)
(using the substitution "u = sin(t)", so du = cos(t) dt)
--------------------------------------------------
integrals II and III:
\(\displaystyle \L \int_0^{2\pi}\, \left[\, 2\sin^2{(t)}\cos{(t)}\, \right]\, =\, 2\, \int_a^b\, u^2\, du\, =\, 0\)
(using the substitution "u = sin(t)", so du = cos(t) dt.)
--------------------------------------------------
Parts I, II, and III all equal zero, so the sum (the entire integral) must be zero.
Is this correct?
