(b) Set up and evaluate \( \iint_{D} e^{y^{3}} d A \) where \( D \) is the region bounded by \( y=\sqrt{x}, y=2 \) and \( x=0 \)

To evaluate the double integral \( \iint_{D} e^{y^{3}} dA \) over the region \( D \) bounded by \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \), we first need to determine the region of integration. 1. **Identify the bounds of the region \( D \)**: - The curve \( y = \sqrt{x} \) can be rewritten as \( x = y^2 \). - The line \( y = 2 \) intersects \( y = \sqrt{x} \) at \( y = 2 \) (since \( \sqrt{x} = 2 \implies x = 4 \)). - The lines \( x = 0 \) (the y-axis) and \( y = 2 \) create our upper boundary. Thus, the region \( D \) is the area under the curve \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 4 \) and above the line \( y = 2 \). 2. **Setting up the integral**: We will integrate with respect to \( y \) first and then with respect to \( x \). The region can be described as: \[ \text{Bounds for } y: 0 \leq y \leq 2 \] For each fixed \( y \), \( x \) ranges from \( 0 \) to \( y^2 \): \[ \text{Bounds for } x: 0 \leq x \leq y^2 \] Thus, we can set up the double integral as: \[ \iint_{D} e^{y^3} \, dA = \int_{0}^{2} \int_{0}^{y^2} e^{y^3} \, dx \, dy \] 3. **Evaluate the inner integral**: \[ \int_{0}^{y^2} e^{y^3} \, dx = e^{y^3} \int_{0}^{y^2} 1 \, dx = e^{y^3} \cdot (y^2 - 0) = y^2 e^{y^3} \] 4. **Evaluate the outer integral**: Now substituting from the inner integral into the outer: \[ \int_{0}^{2} y^2 e^{y^3} \, dy \] To solve this, let's use the substitution \( u = y^3 \), then \( du = 3y^2 \, dy \) or \( dy = \frac{du}{3y^2} \). When \( y = 0 \), \( u = 0 \); when \( y = 2 \), \( u = 8 \). Thus, \[ \int_{0}^{2} y^2 e^{y^3} \, dy = \int_{0}^{8} e^{u} \cdot \frac{1}{3} \, du \] which simplifies to: \[ \frac{1}{3} \int_{0}^{8} e^{u} \, du = \frac{1}{3} [e^{u}]_{0}^{8} = \frac{1}{3} (e^{8} - e^{0}) = \frac{1}{3} (e^{8} - 1) \] Therefore, the value of the original double integral is: \[ \iint_{D} e^{y^3} \, dA = \frac{1}{3} (e^{8} - 1) \]