How do you find the volume of the solid obtained by rotating the region bounded by $y = x$ $y=x$ and $y = x^{2}$ $y=x^2$ about the $x$ $x$ -axis?

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Answer 1 · 2018-04-24T12:27:27

$V = \frac{2 π}{15}$ $V=(2pi)/15$

First we need the points where $x$ $x$ and $x^{2}$ $x^2$ meet.

$x = x^{2}$ $x=x^2$

$x^{x} - x = 0$ $x^x-x=0$

$x(x-1)=0$

$x=0 or 1$

So our bounds are $0$ and $1$ .

When we have two function for the volume, we use:
$V=piint_a^b(f(x)^2-g(x)^2)dx$

$V=piint_0^1(x^2-x^4)dx$

$V=pi[x^3/3-x^5/5]_0^1$

$V=pi(1/3-1/5)=(2pi)/15$

How do you find the volume of the solid obtained by rotating the region bounded by y=xy=xy=x and y=x^2y=x2y=x^2 about the xxx-axis?