To find the volume of the solid obtained by revolving the region enclosed by the curves $y = 3\sqrt{x}$ and $y = 4 - x$ about the x-axis, we can apply the disk method (since we're revolving around a horizontal axis).

Step 1: Find the points of intersection

To determine the bounds of integration, we need to find where the two curves intersect, i.e., where $y = 3\sqrt{x}$ and $y = 4 - x$.

Set $3\sqrt{x} = 4 - x$.

1. Square both sides to eliminate the square root: $9x = (4 - x)^2$ $9x = 16 - 8x + x^2$
2. Rearrange the equation: $x^2 - 17x + 16 = 0$
3. Solve this quadratic equation using the quadratic formula: $x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(1)(16)}}{2(1)}$ $x = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}$ So, $x = 16$ and $x = 1$.

Thus, the curves intersect at $x = 1$ and $x = 16$.

Step 2: Set up the volume integral

The volume of the solid generated by revolving the region about the x-axis is given by the formula:

$V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) dx$ where $R(x)$ is the outer radius (the upper curve), and $r(x)$ is the inner radius (the lower curve). Here:

• $R(x) = 4 - x$ (the upper curve),
• $r(x) = 3\sqrt{x}$ (the lower curve),
• The limits of integration are from $x = 1$ to $x = 16$.

Thus, the volume integral becomes:

$V = \pi \int_{1}^{16} \left[ (4 - x)^2 - (3\sqrt{x})^2 \right] dx$

Step 3: Simplify the integrand

Simplify the integrand:

$(4 - x)^2 = 16 - 8x + x^2$ $(3\sqrt{x})^2 = 9x$

So, the volume integral becomes:

$V = \pi \int_{1}^{16} \left( 16 - 8x + x^2 - 9x \right) dx$ $V = \pi \int_{1}^{16} \left( 16 - 17x + x^2 \right) dx$

Step 4: Integrate

Now, integrate term by term:

$V = \pi \left[ 16x - \frac{17x^2}{2} + \frac{x^3}{3} \right]_{1}^{16}$

Step 5: Evaluate the definite integral

First, evaluate at $x = 16$:

$16(16) - \frac{17(16)^2}{2} + \frac{(16)^3}{3}$ $= 256 - \frac{17(256)}{2} + \frac{4096}{3}$ $= 256 - 2176 + \frac{4096}{3}$ $= 256 - 2176 + 1365.33$ $= -1920 + 1365.33 = -554.67$

Now, evaluate at $x = 1$:

$16(1) - \frac{17(1)^2}{2} + \frac{(1)^3}{3}$ $= 16 - \frac{17}{2} + \frac{1}{3}$ $= 16 - 8.5 + 0.33 = 7.83$

Now subtract the two results:

$V = \pi \left( -554.67 - 7.83 \right) = \pi (-562.5)$

Step 6: Compute the volume

Thus, the volume is:

$V = 562.5\pi \, \text{cubic units}$

Final Answer:

The volume of the solid is $\boxed{562.5\pi}$ cubic units.

Would you like more details or have any further questions?

Related Questions:

1. What is the volume if the region is revolved around the y-axis instead?
2. How does the disk method differ from the shell method in this case?
3. What happens if you revolve the region around a different line, like $y = 1$?
4. How do you find the centroid of the enclosed region?
5. Can you apply the washer method to this problem, and how would the solution change?

Tip:

When solving problems involving volumes of revolution, always visualize the region and the axis of rotation to choose the correct method (disk/washer or shell).