The region is formed by the graphs of y=3sqrt(x) and y=4-x. Find the volume of the solid obtained by revolving the region about the x-axis.

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).

Discover how to calculate the volume of a solid formed by revolving the region between y=3sqrt(x) and y=4-x around the x-axis. This step-by-step solution uses the disk method and involves solving a quadratic equation. Suitable for advanced high school or college-level students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation