Question Completed: 16 of 17 Let $$ R $$ be the region bounded by the following curves. Use the method of your choice to find the volume of the sol generated when $$ R $$ is revolved about the $$ y $$-axis. $$ y=1, y=4 x+1 $$, and $$ x=3 $$ The volume is $$ \square $$. (Type an exact answer.)

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We first sketch the region $$ R $$.

- The line $$ y=1 $$ is horizontal.
- The line $$ y=4x+1 $$ passes through $$(0,1)$$ (since $$4(0)+1=1$$) and $$(3,13)$$ (since $$4(3)+1=13$$).
- The line $$ x=3 $$ is vertical.

So the region $$ R $$ is the triangle with vertices $$(0,1)$$, $$(3,1)$$, and $$(3,13)$$.

Because the solid is generated by revolving $$ R $$ about the $$ y $$-axis, it is convenient to use the cylindrical shells method. At a typical $$ x $$ between $$ 0 $$ and $$ 3 $$:

- The radius of the shell is $$ x $$.
- The height of the shell is the difference between the top and bottom boundaries, i.e. 
  $$
  \text{height} = \big(4x+1\big) - 1 = 4x.
  $$

The thickness of the shell is $$ dx $$.

Thus, the differential volume $$ dV $$ is:
$$
dV = 2\pi \, (\text{radius}) \times (\text{height}) \, dx = 2\pi x (4x) \, dx = 8\pi x^2 \, dx.
$$

Integrate from $$ x=0 $$ to $$ x=3 $$:
$$
V = \int_0^3 8\pi \, x^2 \, dx = 8\pi \int_0^3 x^2 \, dx.
$$

Compute the integral:
$$
\int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} = 9.
$$

Thus, the volume is:
$$
V = 8\pi \cdot 9 = 72\pi.
$$

The exact volume of the solid is $$ \boxed{72\pi} $$.
The volume of the solid is $$ 72\pi $$.

Question Completed: 16 of 17
Let be the region bounded by the following curves. Use the method of your choice to find the volume of the sol
generated when is revolved about the -axis.
, and
The volume is . (Type an exact answer.)

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Question Completed: 16 of 17 Let be the region bounded by the following curves. Use the method of your choice to find the volume of the sol generated when is revolved about the -axis. , and The volume is . (Type an exact answer.)