Area bounded by curve `y^(3) - 9y + x = 0`, and y-axis is

Options

A. (a) `9/2`
B. (b) `9`
C. (c) `(81)/(2)`
D. (d) `81`

Answer

To find the area bounded by the curve \( y^3 - 9y + x = 0 \) and the y-axis, we can follow these steps:

### Step 1: Identify the curve and y-axis
The given curve is \( y^3 - 9y + x = 0 \). The y-axis corresponds to \( x = 0 \).

### Step 2: Find the intersection points
To find the intersection points of the curve with the y-axis, we set \( x = 0 \) in the equation of the curve:
\[ y^3 - 9y + 0 = 0 \implies y^3 - 9y = 0 \]
Factoring out \( y \):
\[ y(y^2 - 9) = 0 \]
This gives us:
\[ y = 0 \quad \text{or} \quad y^2 - 9 = 0 \implies y = \pm 3 \]
Thus, the intersection points are \( (0, 0) \), \( (0, 3) \), and \( (0, -3) \).

### Step 3: Determine the area to be calculated
The area bounded by the curve and the y-axis is between \( y = -3 \) and \( y = 3 \). The curve is symmetric about the x-axis, so we can calculate the area in the first quadrant and then double it.

### Step 4: Rearrange the curve equation
Rearranging the curve equation to express \( x \) in terms of \( y \):
\[ x = 9y - y^3 \]

### Step 5: Set up the integral for area calculation
The area \( A \) can be calculated using the integral:
\[ A = 2 \int_0^3 (9y - y^3) \, dy \]

### Step 6: Calculate the integral
Now we compute the integral:
\[ A = 2 \left[ \int_0^3 9y \, dy - \int_0^3 y^3 \, dy \right] \]

Calculating the first integral:
\[ \int 9y \, dy = \frac{9y^2}{2} \Big|_0^3 = \frac{9 \cdot 3^2}{2} - 0 = \frac{9 \cdot 9}{2} = \frac{81}{2} \]

Calculating the second integral:
\[ \int y^3 \, dy = \frac{y^4}{4} \Big|_0^3 = \frac{3^4}{4} - 0 = \frac{81}{4} \]

### Step 7: Combine the results
Now substituting back into the area calculation:
\[ A = 2 \left( \frac{81}{2} - \frac{81}{4} \right) \]
Finding a common denominator:
\[ A = 2 \left( \frac{162}{4} - \frac{81}{4} \right) = 2 \left( \frac{81}{4} \right) = \frac{162}{4} = \frac{81}{2} \]

### Final Result
Thus, the area bounded by the curve and the y-axis is:
\[ \boxed{\frac{81}{2}} \text{ square units} \]