Volume of the solid generated by revolving the region R enclosed by the curve - Disk and Shell method

Question

Find the volume of the solid generated by revolving the region $R$ enclosed by the curve $y=x^3$, the $x$-axis and the line $x=1$.

(About the line $y=1$ using the disk and shell method.)

Answer 1

Using the Disk Method, $$V=\int_{0}^1\pi[1^2-(1-y)^2]dx=\pi\int_{0}^1[2y-y^2]dx=\pi\int_{0}^1[2x^3-x^6]dx=\dots=\frac{5\pi}{14}$$ and by using the Shell Method, $$V=\int_{0}^1\big[2\pi(1-y)\cdot (1-x)dy\big]=\int_{0}^1\big[2\pi(1-y)\cdot (1-\sqrt[3]y)dy\big]=\dots=\frac{5\pi}{14}$$

Edit: Note that $2\pi(1-y)$ is the circumference of the circle obtained by rotating the point $(x,y)$ from the curve $y=x^3$ about the line $y=1$. Also, $(1-\sqrt[3]y)dy$ is the area of the sliced element of the region.

Answer 2

It seems like you have the disk method down fine, so let's look at the shell method. When describing our shells, we want to find the radius and height of each shell. Notice that we will be integrating along the $y$ values.

When $y = y_0$, for $0 \leq y_0 \leq 1$, the radius of the shell will be $1-y_0$ (the vertical distance from the line $y=1$ to $y_0$). The height of the shell will be $1-y_0^{1/3}$, which is the difference in the $x$ values for the curves $x=1$ and $x^3 = y$ at $y=y_0$.

Then each shell has the volume $2\pi rh \, dy$, or in this case, $2 \pi (1-y_0)(1-y_0^{1/3}) \, dy$.

This gives us the integral $\int\limits_0^1 2 \pi (1-y_0)(1-y_0^{1/3}) \, dy = \frac{5\pi}{14}$.