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=(-1-√(17))/(2)\\ x_2=(-1+√(17))/(2)\\\\ \displaystyle V=\pi\int\limits_0^{(-1+√(17))/(2)}(4-x^2-x)^2\,dx\\ V=\pi\int\limits_0^{(-1+√(17))/(2)}(16-4x^2-4x-4x^2+x^4+x^3-4x+x^3+x^2)\,dx\\ V=\pi\int\limits_0^{(-1+√(17))/(2)}(x^4+2x^3-7x^2-8x+16)\,dx\\ V=\pi \left[(x^5)/(5)+(x^4)/(2)-(7x^3)/(3)-4x^2+16x\right]_0^{(-1+√(17))/(2)}\\`

The rest of solution in the attachment.

There's a mistake in the picture
It shoud be

`V=\pi\left((289√(17)-521)/(60)\right)\approx35`