What is the volume of the solid produced by revolving $f (x) = x^{3}, x \in [0, 3]$ $f(x)=x^3, x in [0,3]$ around $y = - 1$ $y=-1$ ?

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What is the volume of the solid produced by revolving $f (x) = x^{3}, x \in [0, 3]$ $f(x)=x^3, x in [0,3]$ around $y = - 1$ $y=-1$ ?

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Answer 1 · 2018-02-07T10:41:13

See explanation below

Explanation:

We have a revolution solid with a hole inside because the spin axis is y=-1.
Consider a volume element with a width of $d x$ $dx$ and two radios:
$R = 1 + x^{3}$ $R= 1 + x^3$ and $r = 1$ $r=1$ .
So, the volume of this volume element is $d V = π (R^{2} - r^{2}) d x$ $dV=pi(R^2-r^2)dx$
The sbstitution in that formula give us:

$d V = π ({(1 + x^{3})}^{2} - 1^{2}) d x$ $dV=pi((1+x^3)^2-1^2)dx$

Integrating between 0 and 3 wil give us the volume requested

$\int_{0}^{3} π (1 + 2 x^{3} + x^{6} - 1) d x = \int_{0}^{3} π (2 x^{3} + x^{6}) d x$ $int_0^3 pi(1+2x^3+x^6-1)dx=int_0^3pi(2x^3+x^6)dx$

solving this integral we have:

$V = 2 π \frac{x^{4}}{4} + \frac{x^{7}}{7}}_{0}^{3} = 81 \frac{π}{2} + 2187 \frac{π}{7} = \frac{4941}{14} π$ $V=2pix^4/4+x^7/7}_0^3=81pi/2+2187pi/7=4941/14pi$

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What is the volume of the solid produced by revolving $f (x) = x^{3}, x \in [0, 3]$ $f(x)=x^3, x in [0,3]$ around $y = - 1$ $y=-1$ ?

1 Answer

Explanation:

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What is the volume of the solid produced by revolving f(x)=x^3, x in [0,3] f(x)=x3,x∈[0,3]f(x)=x^3, x in [0,3] around y=-1y=−1y=-1?

1 Answer

Explanation:

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What is the volume of the solid produced by revolving $f (x) = x^{3}, x \in [0, 3]$ $f(x)=x^3, x in [0,3]$ around $y = - 1$ $y=-1$ ?