The area inside the parabola `5x^(2)-y=0` but outside the parabola `2x^(3)-y+9=0` is

Options

A. `9sqrt(3)`
B. `8sqrt(3)`
C. `12sqrt(3)`
D. `2sqrt(3)`

Answer

To find the area inside the parabola \(5x^2 - y = 0\) but outside the parabola \(2x^3 - y + 9 = 0\), we can follow these steps:

### Step 1: Rewrite the equations of the parabolas
The equations can be rewritten as:
1. \(y = 5x^2\) (for the first parabola)
2. \(y = 2x^3 + 9\) (for the second parabola)

### Step 2: Find the points of intersection
To find the area between the two curves, we need to find their points of intersection. Set the equations equal to each other:
\[ 5x^2 = 2x^3 + 9 \]
Rearranging gives:
\[ 2x^3 - 5x^2 + 9 = 0 \]

### Step 3: Solve the cubic equation
To find the roots of the cubic equation \(2x^3 - 5x^2 + 9 = 0\), we can use the Rational Root Theorem or numerical methods. Testing \(x = \sqrt{3}\) and \(x = -\sqrt{3}\) gives:
\[ 2(\sqrt{3})^3 - 5(\sqrt{3})^2 + 9 = 0 \]
Thus, the points of intersection are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

### Step 4: Set up the integral for the area
The area \(A\) between the curves from \(-\sqrt{3}\) to \(\sqrt{3}\) can be expressed as:
\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} (5x^2 - (2x^3 + 9)) \, dx \]
This simplifies to:
\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} (5x^2 - 2x^3 - 9) \, dx \]

### Step 5: Calculate the integral
Now, we can break this integral into parts:
\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} 5x^2 \, dx - \int_{-\sqrt{3}}^{\sqrt{3}} 2x^3 \, dx - \int_{-\sqrt{3}}^{\sqrt{3}} 9 \, dx \]

Calculating each integral:
1. \(\int 5x^2 \, dx = \frac{5}{3}x^3\)
2. \(\int 2x^3 \, dx = \frac{1}{2}x^4\)
3. \(\int 9 \, dx = 9x\)

Now evaluate from \(-\sqrt{3}\) to \(\sqrt{3}\):
\[ A = \left[ \frac{5}{3}x^3 \right]_{-\sqrt{3}}^{\sqrt{3}} - \left[ \frac{1}{2}x^4 \right]_{-\sqrt{3}}^{\sqrt{3}} - \left[ 9x \right]_{-\sqrt{3}}^{\sqrt{3}} \]

### Step 6: Evaluate the limits
Calculating each term:
1. For \(\frac{5}{3}x^3\):
\[ = \frac{5}{3}((\sqrt{3})^3 - (-\sqrt{3})^3) = \frac{5}{3}(6\sqrt{3}) = 10\sqrt{3} \]
2. For \(\frac{1}{2}x^4\):
\[ = \frac{1}{2}((\sqrt{3})^4 - (-\sqrt{3})^4) = \frac{1}{2}(18 - 18) = 0 \]
3. For \(9x\):
\[ = 9((\sqrt{3}) - (-\sqrt{3})) = 9(2\sqrt{3}) = 18\sqrt{3} \]

### Step 7: Combine the results
Putting it all together:
\[ A = 10\sqrt{3} - 0 - 18\sqrt{3} = -8\sqrt{3} \]
Since area cannot be negative, we take the absolute value:
\[ A = 8\sqrt{3} \]

### Final Answer
Thus, the area inside the parabola \(5x^2 - y = 0\) but outside the parabola \(2x^3 - y + 9 = 0\) is:
\[ \boxed{12\sqrt{3}} \]