[tex]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}}\\
[/tex]
The rest of solution in the attachment.
There's a mistake in the picture
It shoud be
[tex]V=\pi\left(\dfrac{289\sqrt{17}-521}{60}\right)\approx35[/tex]
