How do you find the volume of the solid obtained by rotating the region bounded by the curves $Y=2x$ and $Y=x^2$ rotated around the x-axis?

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Answer 1 · 2015-05-23T10:05:57

Remembering that the area between $f(x)$ and $g(x)$ , with $f(x)$ over $g(x)$ from $x_1$ to $x_2$ is given by:

$int_(x_1)^(x_2)[f(x)-g(x)]dx$ ,

and since $y=2x$ is over the function $y=x^2$ , and the intercept point is given by the system:

$y=2x$

$y=x^2$

$2x=x^2rArrx(x-2)=0rArr$

$x_1=0$ and $x_2=2$ ,

then:

$int_0^1(2x-x^2)dx=[x^2-x^3/3]_0^2=$

$=[2^2-2^3/3-(0^2-0^3/3)]=4-8/3=4/3$ .

How do you find the volume of the solid obtained by rotating the region bounded by the curves Y=2xY=2x and Y=x^2Y=x^2 rotated around the x-axis?