Find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y=x^3$ and $x$ $x$ -axis in the interval $(1, 2)$ $(1,2)$ ?

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Find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y=x^3$ and $x$ $x$ -axis in the interval $(1, 2)$ $(1,2)$ ?

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Answer 1 · 2017-03-21T05:15:13

Volume is $18 \frac{1}{7} π$ $18 1/7pi$

Explanation:

To find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y=x^3$ , the $x$ $x$ -axis and the lines $x = 1$ $x=1$ and $x = 2$ $x=2$ turn around the $x$ $x$ -axis,

we need to find area of the curve under the curve $y = x^{3}$ $y=x^3$ , between $x = 1$ $x=1$ and $x = 2$ $x=2$ .

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As the same is rotated around $x$ $x$ -axis, we will get the volume of the desired solid.

Hence this volume is $\int_{1}^{2} π {(x^{3})}^{2} d x$ $int_1^2pi(x^3)^2dx$

= $\int_{1}^{2} π x^{6} d x$ $int_1^2pix^6dx$

= $π {[\frac{x^{7}}{7}]}_{1}^{2}$ $pi[x^7/7]_1^2$

= $π {\frac{2^{7}}{7} - \frac{1^{7}}{7}} = \frac{127 π}{7} = 18 \frac{1}{7} π$ $pi{2^7/7-1^7/7}=(127 pi)/7=18 1/7pi$

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Find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y=x^3$ and $x$ $x$ -axis in the interval $(1, 2)$ $(1,2)$ ?

1 Answer

Explanation:

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Find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y=x^3$ and $x$ $x$ -axis in the interval $(1, 2)$ $(1,2)$ ?