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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.]

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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$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

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$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

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