Let's solve the problem step-by-step.

Given:

• The solid region $S$ lies under the surface $z = 4x^2 \sqrt{y}$ and above the rectangle $R = [0, 2] \times [1, 4]$.

(a) Formula for the area of a cross-section in the plane perpendicular to the x-axis

The cross-section at $x$ is obtained by fixing $x$ and integrating $z = 4x^2 \sqrt{y}$ over $y$ from $1$ to $4$.

Formula:

$A(x) = \int_{1}^{4} 4x^2 \sqrt{y} \, dy$

Solve for $A(x)$:

$\int 4x^2 \sqrt{y} \, dy = 4x^2 \int \sqrt{y} \, dy = 4x^2 \cdot \frac{2}{3} y^{3/2} = \frac{8x^2}{3} y^{3/2}$

Evaluate from $y = 1$ to $y = 4$: $A(x) = \frac{8x^2}{3} \left[ (4)^{3/2} - (1)^{3/2} \right]$

Simplify: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8, \quad (1)^{3/2} = 1$ $A(x) = \frac{8x^2}{3} (8 - 1) = \frac{8x^2}{3} \cdot 7 = \frac{56x^2}{3}$

So, the area of the cross-section is: $A(x) = \frac{56x^2}{3}$

(b) Formula for the area of a cross-section in the plane perpendicular to the y-axis

The cross-section at $y$ is obtained by fixing $y$ and integrating $z = 4x^2 \sqrt{y}$ over $x$ from $0$ to $2$.

Formula:

$A(y) = \int_{0}^{2} 4x^2 \sqrt{y} \, dx$

Solve for $A(y)$:

Since $\sqrt{y}$ is constant with respect to $x$: $\int 4x^2 \sqrt{y} \, dx = \sqrt{y} \int 4x^2 \, dx = \sqrt{y} \cdot \frac{4x^3}{3} = \frac{4x^3}{3} \sqrt{y}$

Evaluate from $x = 0$ to $x = 2$: $A(y) = \frac{4}{3} \sqrt{y} \left[ (2)^3 - (0)^3 \right] = \frac{4}{3} \sqrt{y} (8 - 0) = \frac{32}{3} \sqrt{y}$

So, the area of the cross-section is: $A(y) = \frac{32}{3} \sqrt{y}$

(c) Volume of $S$

The volume is the double integral of $z = 4x^2 \sqrt{y}$ over the rectangle $R = [0, 2] \times [1, 4]$.

Formula:

$V = \int_{x=0}^{2} \int_{y=1}^{4} 4x^2 \sqrt{y} \, dy \, dx$

Solve for $V$:

1. Compute the inner integral with respect to $y$: $\int_{1}^{4} 4x^2 \sqrt{y} \, dy = 4x^2 \int_{1}^{4} \sqrt{y} \, dy$

We already computed this in part (a): $\int_{1}^{4} \sqrt{y} \, dy = \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4} = \frac{2}{3} (8 - 1) = \frac{14}{3}$ $\int_{1}^{4} 4x^2 \sqrt{y} \, dy = 4x^2 \cdot \frac{14}{3} = \frac{56x^2}{3}$

2. Compute the outer integral with respect to $x$: $V = \int_{0}^{2} \frac{56x^2}{3} \, dx = \frac{56}{3} \int_{0}^{2} x^2 \, dx$

$\int_{0}^{2} x^2 \, dx = \frac{x^3}{3} \Big|_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$

$V = \frac{56}{3} \cdot \frac{8}{3} = \frac{448}{9}$

So, the volume of $S$ is: $V = \frac{448}{9}$

Final Answers:

(a) $A(x) = \frac{56x^2}{3}$

(b) $A(y) = \frac{32}{3} \sqrt{y}$

(c) $V = \frac{448}{9}$

Would you like me to clarify anything further or provide more details?

Related Questions:

1. What is the geometric interpretation of cross-sections in this problem?
2. How would the formula for $A(x)$ change if the surface equation were different?
3. Can the method used for finding $A(y)$ be applied to other axis orientations?
4. What are some practical applications of computing volumes using double integrals?
5. How can this problem be extended to triple integrals?

Tip:

Always sketch the region of integration to ensure you understand the bounds and the surface orientation clearly.