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Neporo4naja [7]
3 years ago
6

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.]
Mathematics
1 answer:
Alex777 [14]3 years ago
4 0

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}

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