How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y = x^3$ , $x = 0$ $x=0$ , and $x = 1$ $x=1$ rotated around the $y = - 2$ $y=-2$ ?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y = x^3$ , $x = 0$ $x=0$ , and $x = 1$ $x=1$ rotated around the $y = - 2$ $y=-2$ ?

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Answer 1 · 2016-04-18T17:11:51

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The region to be rotated about y=-2 is shown here shaded in red.
Consider an element of length y and width $δ$ $delta$ x at a distance x from the origin. If this this is rotated about y=-2. It would form a circular disc of radius y+2. Its area would be $π {(y + 2)}^{2}$ $pi (y+2)^2$ . Its volume would be $π {(y + 2)}^{2} δ x$ $pi (y+2)^2 delta x$ From this subtract the volume of coaxial region $π 2^{2} δ x$ $pi 2^2 delta x$

Required volume is that of the annular region $π {(y + 2)}^{2} δ x - 4 π δ x$ $pi (y+2)^2 delta x -4pi delta x$

= $π (y^{2} + 4 y) δ x$ $pi(y^2 +4y)delta x$ .

The volume of the solid formed by the whole region would therefore be

$\int_{x = 0}^{1} π (y^{2} + 4 y) d x$ $int_(x=0)^1 pi (y^2 +4y) dx$ . Since y is = $x^{3}$ $x^3$

Integral to solve would be $\int_{x = 0}^{1} π (x^{6} + 4 x^{3}) d x$ $int_(x=0)^1 pi (x^6 +4x^3)dx$

= $π {[\frac{x^{7}}{7} + 4 \frac{x^{4}}{4}]}_{0}^{1}$ $pi[x^7 /7 +4 x^4 /4]_0^1$
= $\frac{8 π}{7}$ $(8pi)/7$

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y = x^3$ , $x = 0$ $x=0$ , and $x = 1$ $x=1$ rotated around the $y = - 2$ $y=-2$ ?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves y = x^3y=x3y = x^3, x=0x=0x=0, and x=1x=1x=1 rotated around the y=-2y=−2y=-2?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = x^{3}$ $y = x^3$ , $x = 0$ $x=0$ , and $x = 1$ $x=1$ rotated around the $y = - 2$ $y=-2$ ?