Question

Find the volume of revolution bounded by the curves y = 4 – x2 , y = x, and x = 0, and is revolved about the vertical axis.

Answer 1

$4-x^2=x\\ x^2+x-4=0\\ \Delta=1^2-4\cdot1\cdot(-4)=1+16=17\\ x_1=\dfrac{-1-\sqrt{17}}{2}\\ x_2=\dfrac{-1+\sqrt{17}}{2}\\\\ \displaystyle V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(4-x^2-x)^2\,dx\\ V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(16-4x^2-4x-4x^2+x^4+x^3-4x+x^3+x^2)\,dx\\ V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(x^4+2x^3-7x^2-8x+16)\,dx\\ V=\pi \left[\dfrac{x^5}{5}+\dfrac{x^4}{2}-\dfrac{7x^3}{3}-4x^2+16x\right]_0^{\dfrac{-1+\sqrt{17}}{2}}$

The rest of solution in the attachment. 

There's a mistake in the picture
It shoud be
$V=\pi\left(\dfrac{289\sqrt{17}-521}{60}\right)\approx35$