Find the volume of the solid by using cylindrical shells?

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Find the volume of the solid by using cylindrical shells?

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Answer 1 · 2018-05-28T06:04:36

$V = 2 π \int_{0}^{1} (y) (\sqrt{y} - y^{3}) \cdot d y = \frac{2 \cdot π}{5} {(u n i t e)}^{3}$ $color(blue)[V=2piint_0^1(y)(sqrty-y^3)*dy=(2*pi)/5 (unite)^3]$

Explanation:

the volume by cylindrical shell method when the curve rotating about the $x-axis$ $"x-axis"$ is given by:

$V = 2 π \int_{a}^{b} (y) (x_{2} - x_{1}) \cdot d y$ $color(red)[V=2piint_a^b(y)(x_2-x_1)*dy]$

$x_{2} = \pm \sqrt{y}$ $x_2=+-sqrty$
but the area enclosed by the curve lies in the first quadrant.

so $x_{2} = \sqrt{y}$ $x_2=sqrty$

$x_{1} = y^{3}$ $x_1=y^3$

lets find the interval of integral.

$\sqrt{y} = y^{3}$ $sqrty=y^3$ square both sides

$y^{6} = y \Rightarrow y^{6} - y = 0$ $y^6=y rArr y^6-y=0$

$y \cdot (y^{5} - 1) = 0$ $y*(y^5-1)=0$

$y = 0 and y = 1$ $y=0 and y=1$

the interval of the integral $y \in [0, 1]$ $y in [0,1]$

now let set up the integral:

$V = 2 π \int_{0}^{1} (y) (\sqrt{y} - y^{3}) \cdot d y$ $V=2piint_0^1(y)(sqrty-y^3)*dy$

$V = 2 π \int_{0}^{1} (y^{\frac{3}{2}} - y^{4}) \cdot d y = {⎡ ⎢ ⎣ - \frac{2 \cdot π \cdot (y^{5} - 2 \cdot y^{\frac{5}{2}})}{5} ⎤ ⎥ ⎦}_{0}^{1}$ $V=2piint_0^1(y^(3/2)-y^4)*dy=[-(2*pi*(y^5-2*y^(5/2)))/5]_0^1$

$= \frac{2 \cdot π}{5} {(u n i t e)}^{3}$ $=(2*pi)/5 (unite)^3$

show below the region revolving (shaded region) :

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Find the volume of the solid by using cylindrical shells?

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