How to evaluate integral volume How to evaluate integral volume
$R$ is the finite region bounded by $x=y$ and $x=4y-y^2$
I do not understand where the area start and end. According to solution manual $V=2\pi\int_0^3...$
Where does $2\pi$ come from? Usually I just use $\pi$.

$x=4y-y^2$ intersects $y=x$ thus
$x=4x-x^2\to x^2-3x=0\to x_1=0;\;x_2=3$
$y=2\pm\sqrt{4-x}$
$$V=\int _{a}^{b}\pi \left(f(x)^{2}-g(x)^{2}\right)dx$$
that is
$$V=\pi\left[\int_0^3\,\left(x^2-(2-\sqrt{4-x})^2\right)\,dx +\int_3^4\,\left((2+\sqrt{4-x})^2-x^2\right)\,dx\right]$$
$$V=\pi\left[\int_0^3\,\left(x^2+x+4 \sqrt{4-x}-8\right)\,dx +\int_3^4\,\left(x^2+x-4 \sqrt{4-x}-8\right)\,dx\right]$$
$$V=\pi\left(\left\vert \frac{x^3}{3}+\frac{x^2}{2}+\frac{1}{3} (-8) (4-x)^{3/2}-8 x\right\vert_0^3+\left\vert \frac{x^3}{3}+\frac{x^2}{2}+\frac{8}{3} (4-x)^{3/2}+8 (4-x)\right\vert_3^4\right)=\\V=3\pi$$
Hope this helps
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