If the region bounded by the lines `y=-4/3x+4` , x=0 , and y=0 is rotated about the y-axis, the volume of the figure formed is

Options

A. 18.8
B. 37.7
C. 56.5
D. 84.8

Explanation

To find the volume of the solid formed by rotating the region bounded by the lines \( y = -\frac{4}{3}x + 4 \), \( x = 0 \), and \( y = 0 \) about the y-axis, we can follow these steps:

### Step 1: Determine the points of intersection
First, we need to find the points where the line intersects the axes.

1. **Find the x-intercept**:
Set \( y = 0 \) in the equation \( y = -\frac{4}{3}x + 4 \):
\[ 0 = -\frac{4}{3}x + 4 \implies \frac{4}{3}x = 4 \implies x = 3 \]
So, the x-intercept is \( (3, 0) \).

2. **Find the y-intercept**:
Set \( x = 0 \) in the equation:
\[ y = -\frac{4}{3}(0) + 4 \implies y = 4 \]
So, the y-intercept is \( (0, 4) \).

### Step 2: Sketch the region
The region bounded by the lines \( y = -\frac{4}{3}x + 4 \), \( x = 0 \) (the y-axis), and \( y = 0 \) (the x-axis) is a triangle with vertices at \( (0, 0) \), \( (3, 0) \), and \( (0, 4) \).

### Step 3: Set up the volume integral
To find the volume of the solid formed by rotating this region about the y-axis, we can use the shell method. The formula for the volume \( V \) using the shell method is:
\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]
where \( f(x) \) is the function representing the height of the shell and \( a \) and \( b \) are the bounds of integration.

From the line equation, we can express \( x \) in terms of \( y \):
\[ y = -\frac{4}{3}x + 4 \implies x = \frac{3}{4}(4 - y) \]

### Step 4: Determine the limits of integration
The limits of integration for \( y \) are from \( 0 \) to \( 4 \).

### Step 5: Write the integral
Now we can set up the integral:
\[ V = 2\pi \int_{0}^{4} \left(\frac{3}{4}(4 - y)\right) y \, dy \]

### Step 6: Simplify and compute the integral
First, simplify the integrand:
\[ = 2\pi \int_{0}^{4} \left(\frac{3}{4}(4y - y^2)\right) \, dy = \frac{3\pi}{2} \int_{0}^{4} (4y - y^2) \, dy \]

Now compute the integral:
\[ = \frac{3\pi}{2} \left[ 2y^2 - \frac{y^3}{3} \right]_{0}^{4} \]
Calculating at the bounds:
\[ = \frac{3\pi}{2} \left[ 2(4^2) - \frac{(4^3)}{3} \right] = \frac{3\pi}{2} \left[ 32 - \frac{64}{3} \right] = \frac{3\pi}{2} \left[ \frac{96}{3} - \frac{64}{3} \right] = \frac{3\pi}{2} \left[ \frac{32}{3} \right] = \frac{32\pi}{2} = 16\pi \]

### Final Answer
Thus, the volume of the solid formed is:
\[ V = 16\pi \]