How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

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How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

y=x $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$

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Answer 1 · 2018-05-13T20:34:24

We must imagine that the graph has radius $r$ $r$ . Then the volume will be $π r^{2} h$ $pir^2h$ . In this case we have:

$V = π \int_{0}^{1} {(x \sqrt{1 - x^{2}})}^{2} d x$ $V = pi int_0^1 (xsqrt(1 - x^2))^2dx$

Therefore

$V = π \int_{0}^{1} x^{2} (1 - x^{2}) d x$ $V = pi int_0^1 x^2(1 - x^2)dx$

$V = π \int_{0}^{1} x^{2} - x^{4} d x$ $V= pi int_0^1 x^2 - x^4 dx$

$V = π {[\frac{1}{3} x^{3} - \frac{1}{5} x^{5}]}_{0}^{3}$ $V = pi[1/3x^3 - 1/5x^5]_0^1$

$V = π (\frac{1}{3} - \frac{1}{5})$ $V = pi(1/3 - 1/5)$

$V = \frac{2}{15} π$ $V = 2/15pi$ cubic units

Hopefully this helps!

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How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

y=x $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$

1 Answer

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How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

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y=x $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$