The curve `x = 4 - 3y - y^(2)` cuts the y-axis into two points P and Q. then the area enclosed by the y-axis and the portion of the curve which lies between P and Q is

Options

A. 20 sq. units
B. 18 sq. units
C. 17 sq. units
D. none of these

Answer

To find the area enclosed by the y-axis and the portion of the curve \( x = 4 - 3y - y^2 \) between the points where it intersects the y-axis, we will follow these steps:

### Step 1: Find the points of intersection with the y-axis
To find the points where the curve intersects the y-axis, we set \( x = 0 \).

\[ 0 = 4 - 3y - y^2 \]

Rearranging gives:

\[ y^2 + 3y - 4 = 0 \]

### Step 2: Solve the quadratic equation
We will use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 3, c = -4 \).

\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ y = \frac{-3 \pm \sqrt{9 + 16}}{2} \]
\[ y = \frac{-3 \pm \sqrt{25}}{2} \]
\[ y = \frac{-3 \pm 5}{2} \]

Calculating the two possible values:

1. \( y = \frac{2}{2} = 1 \)
2. \( y = \frac{-8}{2} = -4 \)

Thus, the points of intersection are \( P(0, 1) \) and \( Q(0, -4) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curve and the y-axis from \( y = -4 \) to \( y = 1 \) can be expressed as:

\[ A = \int_{-4}^{1} (4 - 3y - y^2) \, dy \]

### Step 4: Evaluate the integral
Now we will compute the integral:

\[ A = \int_{-4}^{1} (4 - 3y - y^2) \, dy \]

Calculating the integral term by term:

\[ = \left[ 4y - \frac{3y^2}{2} - \frac{y^3}{3} \right]_{-4}^{1} \]

Now we will evaluate this from \( -4 \) to \( 1 \):

1. **Upper limit \( y = 1 \)**:
\[ 4(1) - \frac{3(1)^2}{2} - \frac{(1)^3}{3} = 4 - \frac{3}{2} - \frac{1}{3} \]
Converting to a common denominator (6):
\[ = \frac{24}{6} - \frac{9}{6} - \frac{2}{6} = \frac{24 - 9 - 2}{6} = \frac{13}{6} \]

2. **Lower limit \( y = -4 \)**:
\[ 4(-4) - \frac{3(-4)^2}{2} - \frac{(-4)^3}{3} = -16 - \frac{48}{2} + \frac{64}{3} \]
Simplifying:
\[ = -16 - 24 + \frac{64}{3} = -40 + \frac{64}{3} \]
Converting to a common denominator (3):
\[ = -\frac{120}{3} + \frac{64}{3} = -\frac{56}{3} \]

### Step 5: Combine the results
Now we combine the results from the upper and lower limits:

\[ A = \left( \frac{13}{6} - \left(-\frac{56}{3}\right) \right) = \frac{13}{6} + \frac{112}{6} = \frac{125}{6} \]

Thus, the area enclosed by the y-axis and the curve between points P and Q is:

\[ \boxed{\frac{125}{6}} \]