Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated line.
-About the line $y = - 1$

A) $\frac { 2048 } { 15 } \pi$
B) $\frac { 7168 } { 15 } \pi$
C) $\frac { 2368 } { 15 } \pi$
D) $\frac { 1184 } { 15 } \pi$

Question

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated line.
-About the line $y = - 1$

A) $\frac { 2048 } { 15 } \pi$ 
B) $\frac { 7168 } { 15 } \pi$ 
C) $\frac { 2368 } { 15 } \pi$ 
D) $\frac { 1184 } { 15 } \pi$

Quizplus · Accepted Answer

The region to be revolved is bounded by the curves $y=x^2-4x+5,\ y=0,$ and $x=1$.

We can see that the axis of rotation is parallel to the direction of integration (in this case, vertical), so we will be using the shell method.
We want to integrate along the $x$-axis. Thus, we will be integrating with respect to $x$.

For each $x$, the corresponding cylinder has radius $x+1$ (distance from $y=-1$ to curve) and height $x^2-4x+5$. Thus, its volume is $$V(x) = 2\pi(x+1)\cdot (x^2-4x+5)$$

Now, we want to evaluate the integral $$\int_{0}^{1} V(x) dx$$

\begin{align*} \int_{0}^{1}& 2\pi(x+1)\cdot(x^2-4x+5)\ dx \ &= 2\pi\int_{0}^{1} (x^3-2x^2-3x+5)\ dx \ &= 2\pi\left[\frac{x^4}{4}-\frac23x^3-\frac32x^2+5xight]_{0}^{1} \ &= \frac{2\pi}{3}\left(4-8-6+20ight) \ &= \boxed{\frac{2368}{15}\pi} \end{align*}

Use the Shell Method to Find the Volume of the Solid