So I have the area enclosed by $y = 0.5x^3 + 4, y = 2, x = 2$ and the y-axis. Suppose this was rotated around the x-axis and the washer method is used to find the volume. Then $V = π∫ (y^2(+) - y^2(-))dx$
Now based on the area the are being rotated on the x-axis is between $x = 0$ and $x = 2$, whilst the y-axis is between $y = 2$ and $y = 8$. I assume since the rotation is around the x-axis that we integrate with the limits of 0 and 2. However, what are the values for the rest of the formula?
V = $\pi \int_0^2( (0.5x³+4)²-4)\,dx\doteq 140.025$
https://www.wolframalpha.com/input/?i=%5Cpi+%5Cint_0%5E2%28+%280.5x%C2%B3%2B4%29%C2%B2-4%29%5C%2Cdx
If I understood the question correctly, I would say that
V = $2\pi \int_2^4 2y\,dy + 2\pi \int_4^8 y\cdot (2-\sqrt[3]{2(y-4)})\,dy \doteq 140.025$
https://www.wolframalpha.com/input/?i=2%5Cpi+%5Cint_2%5E4+2y%5C%2Cdy+%2B+2%5Cpi+%5Cint_4%5E8+y%5Ccdot+%282-%5Csqrt%5B3%5D%7B2%28y-4%29%7D%29%5C%2Cdy
or
V = $2\pi \int_2^8 2y\,dy - 2\pi \int_4^8 y\cdot\sqrt[3]{2(y-4)}\,dy \doteq 140.025$
https://www.wolframalpha.com/input/?i=2%5Cpi+%5Cint_2%5E8+2y%5C%2Cdy+-+2%5Cpi+%5Cint_4%5E8+y%5Ccdot%5Csqrt%5B3%5D%7B2%28y-4%29%7D%5C%2Cdy