Question

Use Wolframalpha to find the exact volume of the solid obtained by rotating about x = [pi]/2 the region bounded by y = sin^2x, y = sin^4x, x in [0,[pi]/2]. [Notation: c in [a,b] means a leq c leq b.]

Answer 1

Answer:

Step-by-step explanation:

Here the region between two curves is rotated about a vertical line.

The functions are

$y = sin^2x, \\y = sin^4x, \\x in [0,[pi]/2].$

Intersecting points are x=0 and x =pi/2

Since rotated about x = pi/2 we get

using cylindrical shell method

Volume = $2\pi rh\\=2\pi \int\limits^\frac{\pi}{2} _0 {xy} \, dx \\=2\pi \int\limits^\frac{\pi}{2} _0 (x+\frac{\pi}{2} )(sin^2 x -sin^4 x) dx$

From wolfram alpha we find that

Volume= $2\pi (\frac{3\pi^2 }{64)} =\frac{6\pi^3}{64}$