Compute the length of the path OABCO consisting of portions of the curves `y^(2)= 2x^(3) and x^(2) + y^(2)= 20`

Answer

To compute the length of the path OABCO consisting of portions of the curves \(y^2 = 2x^3\) and \(x^2 + y^2 = 20\), we will follow these steps:

### Step 1: Identify the curves and their intersections
The first curve is \(y^2 = 2x^3\) and the second curve is \(x^2 + y^2 = 20\). We need to find the points where these two curves intersect.

Substituting \(y^2 = 2x^3\) into the second equation:
\[ x^2 + 2x^3 = 20 \]
Rearranging gives:
\[ 2x^3 + x^2 - 20 = 0 \]

### Step 2: Solve for x
We can use trial and error or synthetic division to find the roots of the polynomial. Testing \(x = 2\):
\[ 2(2)^3 + (2)^2 - 20 = 16 + 4 - 20 = 0 \]
Thus, \(x = 2\) is a root. We can factor the polynomial as:
\[ (x - 2)(2x^2 + 5x + 10) = 0 \]
The quadratic does not yield real roots (discriminant \(5^2 - 4 \cdot 2 \cdot 10 < 0\)), so the only intersection point is at \(x = 2\).

### Step 3: Find the corresponding y-coordinate
Substituting \(x = 2\) back into \(y^2 = 2x^3\):
\[ y^2 = 2(2^3) = 16 \implies y = 4 \text{ or } y = -4 \]
Thus, the intersection points are \(A(2, 4)\) and \(B(2, -4)\).

### Step 4: Calculate the length of OA and OC
The path from \(O(0, 0)\) to \(A(2, 4)\) can be calculated using the arc length formula:
\[ L = \int_{0}^{2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
To find \(\frac{dy}{dx}\), we differentiate \(y^2 = 2x^3\):
\[ 2y \frac{dy}{dx} = 6x^2 \implies \frac{dy}{dx} = \frac{3x^2}{y} \]
Substituting \(y = \sqrt{2x^3}\):
\[ \frac{dy}{dx} = \frac{3x^2}{\sqrt{2x^3}} = \frac{3\sqrt{2}x^{1/2}}{2} \]

### Step 5: Compute the integral for OA
Now substituting \(\frac{dy}{dx}\) into the arc length formula:
\[ L_{OA} = \int_{0}^{2} \sqrt{1 + \left(\frac{3\sqrt{2}x^{1/2}}{2}\right)^2} \, dx \]
Calculating the integral:
\[ = \int_{0}^{2} \sqrt{1 + \frac{9}{4} \cdot 2x} \, dx = \int_{0}^{2} \sqrt{1 + \frac{9}{2}x} \, dx \]

### Step 6: Evaluate the integral
Let \(u = 1 + \frac{9}{2}x\), then \(du = \frac{9}{2}dx\), so \(dx = \frac{2}{9}du\).
Changing the limits accordingly:
- When \(x = 0\), \(u = 1\)
- When \(x = 2\), \(u = 10\)

Thus, the integral becomes:
\[ L_{OA} = \frac{2}{9} \int_{1}^{10} \sqrt{u} \, du = \frac{2}{9} \cdot \frac{2}{3} \left[u^{3/2}\right]_{1}^{10} = \frac{4}{27} \left[10^{3/2} - 1^{3/2}\right] \]

### Step 7: Calculate the length of AB and BC
The length \(AB + BC\) can be calculated as follows:
Since \(AB\) and \(BC\) are both segments of the circle \(x^2 + y^2 = 20\), the length can be computed from \(A(2, 4)\) to \(B(2, -4)\) and back to \(O\).

The length of the arc \(AB\) can be calculated using the formula for the circumference of a circle and the angle subtended by the arc.

### Final Step: Sum the lengths
The total length of the path \(OABCO\) is:
\[ L_{OABCO} = 2L_{OA} + 2L_{AB} \]