Find the volume of the region bounded by $y-x^3+1$, $y=0$, $x=0$ and $x=1$ revolved around the $x$-axis.
I got $\frac{9\pi}{14}$ but the actual answer was $\frac{23\pi}{14}$.
My work was to first get y by itself which made y= x^3 - 1
Then I created the equation $\int _0^1\pi \left|x^3-1\right|^2dx\:$
And when solved I get $\frac{9\pi}{14}$ which is wrong
It is immediately obvious what we should do here. Washer integration(disk technically, but whatever)
Outer function is $y=x^3+1$
Inner function is $y=0$
Hence, $$V=\int_0^1\pi\left((x^3+1)^2-(0)^2\right)\text{ d}x$$ We do not need absolute values because our function is always positive on this region of integration.
We have \begin{align*} V&=\pi\int_0^1(x^3+1)^2-(0)^2\text{ d}x\\ &=\pi\int_0^1 x^6+2x^3+1\text{ d}x\\ &=\left.\pi\left(\frac{x^7}7+\frac{x^4}2+x\right)\right|^1_0\\ &=\pi\left[\left(\frac{1^7}7+\frac{1^4}2+1\right)-\left(\frac{0^7}7+\frac{0^4}2+0\right)\right]\\ &=\pi\left(\frac{1}7+\frac{1}2+1\right)=\frac{23\pi}{14} \end{align*}