Find the area bounded by the y-axis and the curve `x = e^(y) sin piy` between y = 0, y = 1.

Answer

To find the area bounded by the y-axis and the curve \( x = e^y \sin(\pi y) \) between \( y = 0 \) and \( y = 1 \), we will follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) can be expressed as:
\[ A = \int_{0}^{1} e^y \sin(\pi y) \, dy \]

### Step 2: Apply integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let's choose:
- \( u = \sin(\pi y) \) (thus \( du = \pi \cos(\pi y) \, dy \))
- \( dv = e^y \, dy \) (thus \( v = e^y \))

Now we can apply the integration by parts:
\[ A = \left[ \sin(\pi y) e^y \right]_{0}^{1} - \int_{0}^{1} e^y \pi \cos(\pi y) \, dy \]

### Step 3: Evaluate the boundary term
Calculating the boundary term:
\[ \left[ \sin(\pi y) e^y \right]_{0}^{1} = \sin(\pi \cdot 1) e^1 - \sin(\pi \cdot 0) e^0 = 0 \cdot e - 0 \cdot 1 = 0 \]
Thus, the area simplifies to:
\[ A = -\pi \int_{0}^{1} e^y \cos(\pi y) \, dy \]

### Step 4: Apply integration by parts again
We will apply integration by parts again on the integral \( \int e^y \cos(\pi y) \, dy \). Let:
- \( u = \cos(\pi y) \) (thus \( du = -\pi \sin(\pi y) \, dy \))
- \( dv = e^y \, dy \) (thus \( v = e^y \))

Now we apply integration by parts:
\[ \int e^y \cos(\pi y) \, dy = \left[ \cos(\pi y) e^y \right]_{0}^{1} - \int_{0}^{1} e^y (-\pi \sin(\pi y)) \, dy \]
This gives:
\[ \int e^y \cos(\pi y) \, dy = \left[ \cos(\pi y) e^y \right]_{0}^{1} + \pi \int_{0}^{1} e^y \sin(\pi y) \, dy \]

### Step 5: Evaluate the boundary term for the cosine integral
Calculating the boundary term:
\[ \left[ \cos(\pi y) e^y \right]_{0}^{1} = \cos(\pi \cdot 1) e^1 - \cos(\pi \cdot 0) e^0 = -1 \cdot e - 1 \cdot 1 = -e - 1 \]
Thus, we have:
\[ \int_{0}^{1} e^y \cos(\pi y) \, dy = (-e - 1) + \pi A \]

### Step 6: Substitute back into the area equation
Substituting back into the area equation:
\[ A = -\pi \left( -e - 1 + \pi A \right) \]
This simplifies to:
\[ A = \pi (e + 1) - \pi^2 A \]

### Step 7: Solve for A
Rearranging gives:
\[ A + \pi^2 A = \pi (e + 1) \]
\[ A (1 + \pi^2) = \pi (e + 1) \]
Thus,
\[ A = \frac{\pi (e + 1)}{1 + \pi^2} \]

### Final Answer
The area bounded by the y-axis and the curve \( x = e^y \sin(\pi y) \) between \( y = 0 \) and \( y = 1 \) is:
\[ A = \frac{\pi (e + 1)}{1 + \pi^2} \]