The area between the curves `x=4y-y^(2)` and `0` is `lambda` square units, then the value of `3lambda` is equal to

Options

A. 28
B. 30
C. 32
D. 36

Answer

To find the area between the curves \( x = 4y - y^2 \) and \( x = 0 \), we will follow these steps:

### Step 1: Identify the curves and their intersection points
The curve \( x = 4y - y^2 \) is a downward-opening parabola. To find the intersection points with the line \( x = 0 \), we set:
\[ 4y - y^2 = 0 \]
Factoring gives:
\[ y(4 - y) = 0 \]
Thus, the solutions are:
\[ y = 0 \quad \text{and} \quad y = 4 \]
So, the points of intersection are \( (0, 0) \) and \( (0, 4) \).

### Step 2: Set up the integral for the area
The area \( A \) between the curve and the line \( x = 0 \) can be found by integrating the function \( x = 4y - y^2 \) with respect to \( y \) from \( y = 0 \) to \( y = 4 \):
\[ A = \int_{0}^{4} (4y - y^2) \, dy \]

### Step 3: Calculate the integral
Now, we compute the integral:
\[ A = \int_{0}^{4} (4y - y^2) \, dy = \int_{0}^{4} 4y \, dy - \int_{0}^{4} y^2 \, dy \]
Calculating each integral separately:
1. For \( \int 4y \, dy \):
\[ \int 4y \, dy = 2y^2 \bigg|_{0}^{4} = 2(4^2) - 2(0^2) = 32 \]
2. For \( \int y^2 \, dy \):
\[ \int y^2 \, dy = \frac{y^3}{3} \bigg|_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} \]

Now substituting back into the area calculation:
\[ A = 32 - \frac{64}{3} \]
To combine these, we convert \( 32 \) into a fraction:
\[ 32 = \frac{96}{3} \]
Thus,
\[ A = \frac{96}{3} - \frac{64}{3} = \frac{32}{3} \]

### Step 4: Find \( \lambda \) and calculate \( 3\lambda \)
Since we have found that the area \( A = \lambda = \frac{32}{3} \), we need to find \( 3\lambda \):
\[ 3\lambda = 3 \times \frac{32}{3} = 32 \]

### Final Answer
Thus, the value of \( 3\lambda \) is \( \boxed{32} \).

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