Find the area of the region bounded by the curve `x = at^(2)` and `y = 2at` between the ordinates corresponding to t = 1 and t = 2.

Answer

To find the area of the region bounded by the curves \( x = at^2 \) and \( y = 2at \) between the ordinates corresponding to \( t = 1 \) and \( t = 2 \), we will follow these steps:

### Step 1: Identify the limits of integration
The parameter \( t \) varies from \( t = 1 \) to \( t = 2 \). We will find the corresponding \( x \) values for these \( t \) values.

- For \( t = 1 \):
\[ x_1 = a(1^2) = a \]
- For \( t = 2 \):
\[ x_2 = a(2^2) = 4a \]

Thus, the limits of integration will be from \( x = a \) to \( x = 4a \).

### Step 2: Express \( y \) in terms of \( x \)
We have the parametric equations:
- \( x = at^2 \)
- \( y = 2at \)

From the equation \( x = at^2 \), we can express \( t \) in terms of \( x \):
\[ t = \sqrt{\frac{x}{a}} \]

Now substitute this \( t \) into the equation for \( y \):
\[ y = 2at = 2a\sqrt{\frac{x}{a}} = 2\sqrt{ax} \]

### Step 3: Set up the integral for the area
The area \( A \) under the curve from \( x = a \) to \( x = 4a \) is given by the integral:
\[ A = \int_{a}^{4a} y \, dx = \int_{a}^{4a} 2\sqrt{ax} \, dx \]

### Step 4: Calculate the integral
First, factor out the constant \( 2\sqrt{a} \):
\[ A = 2\sqrt{a} \int_{a}^{4a} \sqrt{x} \, dx \]

Now, we can compute the integral:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]

Evaluating the definite integral:
\[ A = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_{a}^{4a} \]

Calculating the limits:
\[ = 2\sqrt{a} \left( \frac{2}{3} (4a)^{3/2} - \frac{2}{3} (a)^{3/2} \right) \]
\[ = 2\sqrt{a} \left( \frac{2}{3} (8a^{3/2}) - \frac{2}{3} (a^{3/2}) \right) \]
\[ = 2\sqrt{a} \left( \frac{2}{3} (8a^{3/2} - a^{3/2}) \right) \]
\[ = 2\sqrt{a} \left( \frac{2}{3} (7a^{3/2}) \right) \]
\[ = \frac{28}{3} a^2 \]

### Step 5: Final area
Thus, the area of the region bounded by the curves is:
\[ \boxed{\frac{28}{3} a^2} \]