Solution volume of solid of revolution

Question

Find the solid volume formed by the rotation of the region bounded by $x = 0, x = 1, y = 0, y = 5 + x ^ 6$ on the $x-\text{axis}.$

I make: $\int_0^1 \pi \cdot ((5+x^6)-0)^2$. And find: $(\cfrac{1}{13} +\cfrac{10}{7} + 25) \cdot \pi.$

Answer 1

Yes you are correct.

The volume formed by the rotation of the region bounded by $x=0, x=1,y=0$ and the curve $y=5+x^6$ on the $x$-axis is $$V=\int_{0}^{1}\pi y^2dx$$

Thus we have $$V=\pi\int_{0}^{1}(5+x^6)^2dx$$ $$=\pi(25+\frac{1}{13}+\frac{10}{7})$$ $$=\frac{2412\pi}{91} \space\text{unit$^{3}$}.$$