Compute the area of the region enclosed by the curve `x= a sin t, y= b sin 2t`

Answer

To compute the area of the region enclosed by the parametric curve given by \( x = a \sin t \) and \( y = b \sin 2t \), we will use the formula for the area enclosed by a parametric curve, which is given by:

\[ A = \int_{t_1}^{t_2} y \frac{dx}{dt} dt \]

### Step 1: Find \(\frac{dx}{dt}\)

Given \( x = a \sin t \), we differentiate \( x \) with respect to \( t \):

\[ \frac{dx}{dt} = a \cos t \]

### Step 2: Substitute \( y \) and \(\frac{dx}{dt}\) into the area formula

Now, substitute \( y = b \sin 2t \) and \(\frac{dx}{dt}\) into the area formula:

\[ A = \int_{t_1}^{t_2} b \sin 2t \cdot a \cos t \, dt \]

This simplifies to:

\[ A = ab \int_{t_1}^{t_2} \sin 2t \cos t \, dt \]

### Step 3: Use the identity for \(\sin 2t\)

Recall that \(\sin 2t = 2 \sin t \cos t\). Thus, we can rewrite the integral as:

\[ A = ab \int_{t_1}^{t_2} 2 \sin t \cos t \cos t \, dt = 2ab \int_{t_1}^{t_2} \sin t \cos^2 t \, dt \]

### Step 4: Determine the limits of integration

To find the limits \( t_1 \) and \( t_2 \), we need to determine the period of the curve. The function \( y = b \sin 2t \) has a period of \( \pi \), and since \( x = a \sin t \) has a period of \( 2\pi \), we can take \( t_1 = 0 \) and \( t_2 = \pi \) for one complete traversal of the curve.

### Step 5: Evaluate the integral

Now we evaluate the integral:

\[ A = 2ab \int_{0}^{\pi} \sin t \cos^2 t \, dt \]

Using the identity \(\cos^2 t = \frac{1 + \cos 2t}{2}\):

\[ A = 2ab \int_{0}^{\pi} \sin t \left(\frac{1 + \cos 2t}{2}\right) dt = ab \int_{0}^{\pi} \sin t (1 + \cos 2t) dt \]

This can be split into two integrals:

\[ A = ab \left( \int_{0}^{\pi} \sin t \, dt + \int_{0}^{\pi} \sin t \cos 2t \, dt \right) \]

### Step 6: Compute the integrals

1. The first integral:

\[ \int_{0}^{\pi} \sin t \, dt = [-\cos t]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 2 \]

2. The second integral can be evaluated using integration by parts or known results:

\[ \int_{0}^{\pi} \sin t \cos 2t \, dt = 0 \]

This is because \(\sin t \cos 2t\) is an odd function over the interval \([0, \pi]\).

### Step 7: Combine results

Thus, we have:

\[ A = ab (2 + 0) = 2ab \]

### Final Result

Therefore, the area of the region enclosed by the curve is:

\[ \boxed{2ab} \]