help with calculus 2 volume problem?

Your teacher is an IDIOT. The region is not "rotated about the x-axis." The region is the base of a solid whose cross-sections are triangles perpendicular to the x-axis. I am sorry you have to be taught by a MORON.

OK, it's not hard to write the integral. Slice the object perpendicular to the x-axis, so that the thickness of each triangular slab is dx. The base of each triangle is sin(x), and its height will be
[sin(x)*sqrt(3)/2], so its area will be
[sqrt(3)/4]*sin^2(x).
Then integrate from x = 0 to pi.

The indefinite integral of sin^2(x) is
(1/2)[x - (1/2)sin(2x)],
easy to verify as the derivative is
(1/2)[1 - cos(2x)]
= (1/2) - (1/2)cos^2(x) + (1/2)sin^2(x)
= (1/2)sin^2(x) + (1/2)sin^2(x) = sin^2(x).

So plug in the limits x = pi and 0 in the expression
[sqrt(3)/4]*(1/2)[x - (1/2)sin(2x)], obtaining
sqrt(3)/8 = about 0.2165.

Your teacher is a DUNCE.

Since the cross sections are perpendicular to the x axis, the representative 'length' must be drawn vertically.
Length = y = sin x

Area of an equilateral triangle = (√3/4) s^2 , where s is the length
= (√3/4) (sin x)^2

.... .... ... ... ... ..pi
Volume = (√3/4) ∫ (sin x)^2 dx
.... .... .... ... ....0

∫ (sin x)^2 dx = ∫ (1-cos 2x) / 2 dx
= (1/2) x - (1/2)(1/2) sin 2x
= (1/2) x - (1/4) sin 2x

Let F(x) = (1/2) x - (1/4) sin 2x

substitute the upper limit pi
F(pi) = pi/2 - 0

substitute the lower limit 0
F(0) = 0

subtract:
F(pi)-F(0) = pi/2

Volume = (√3/4) (pi/2)
= (√3 /8)pi