To find the area enclosed by the curves \( x^2 = 4by \) and \( y = mx \), we can follow these steps:
### Step 1: Identify the curves
The first curve is given by the equation \( x^2 = 4by \), which is a parabola that opens upwards. The second curve is a straight line given by \( y = mx \), where \( m \) is the slope.
### Step 2: Find the points of intersection
To find the points where these two curves intersect, we substitute \( y = mx \) into the equation of the parabola:
\[
x^2 = 4b(mx)
\]
This simplifies to:
\[
x^2 - 4bmx = 0
\]
Factoring out \( x \):
\[
x(x - 4bm) = 0
\]
Thus, the solutions for \( x \) are:
\[
x = 0 \quad \text{or} \quad x = 4bm
\]
Now, we can find the corresponding \( y \) values. For \( x = 0 \):
\[
y = m \cdot 0 = 0
\]
For \( x = 4bm \):
\[
y = m(4bm) = 4bm^2
\]
So the points of intersection are \( (0, 0) \) and \( (4bm, 4bm^2) \).
### Step 3: Calculate the area of the triangle formed by the points
The area \( A \) of the triangle formed by the points \( (0, 0) \), \( (4bm, 0) \), and \( (4bm, 4bm^2) \) can be calculated using the formula for the area of a triangle:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is \( 4bm \) and the height is \( 4bm^2 \):
\[
A = \frac{1}{2} \times 4bm \times 4bm^2 = \frac{1}{2} \times 16b^2m^3 = 8b^2m^3
\]
### Step 4: Calculate the area under the parabola
Next, we need to find the area under the parabola from \( x = 0 \) to \( x = 4bm \). The equation of the parabola can be rewritten as:
\[
y = \frac{x^2}{4b}
\]
The area \( A_p \) under the parabola is given by the integral:
\[
A_p = \int_{0}^{4bm} \frac{x^2}{4b} \, dx
\]
Calculating this integral:
\[
A_p = \frac{1}{4b} \int_{0}^{4bm} x^2 \, dx
\]
The integral of \( x^2 \) is:
\[
\int x^2 \, dx = \frac{x^3}{3}
\]
Thus,
\[
A_p = \frac{1}{4b} \left[ \frac{x^3}{3} \right]_{0}^{4bm} = \frac{1}{4b} \left( \frac{(4bm)^3}{3} - 0 \right)
\]
Calculating \( (4bm)^3 \):
\[
(4bm)^3 = 64b^3m^3
\]
So,
\[
A_p = \frac{1}{4b} \cdot \frac{64b^3m^3}{3} = \frac{64b^2m^3}{12} = \frac{16b^2m^3}{3}
\]
### Step 5: Find the enclosed area
The area enclosed by the two curves is the area of the triangle minus the area under the parabola:
\[
A_{enclosed} = A - A_p = 8b^2m^3 - \frac{16b^2m^3}{3}
\]
To combine these, we need a common denominator:
\[
A_{enclosed} = \frac{24b^2m^3}{3} - \frac{16b^2m^3}{3} = \frac{8b^2m^3}{3}
\]
### Final Result
Thus, the area enclosed by the curves \( x^2 = 4by \) and \( y = mx \) is:
\[
\boxed{\frac{8}{3} b^2 m^3} \text{ square units.}
\]
