Thank you for sharing the images. From the uploaded content, it seems you are working on a calculus problem involving a solid region $S$ under the surface $z = 2x^2 \sqrt{y}$ over a rectangle in the $xy$-plane, and you're calculating areas of cross-sections and the volume of $S$.

Let’s break it down:

(a) Cross-sections perpendicular to the $x$-axis

The given surface equation is $z = 2x^2 \sqrt{y}$, and you are tasked with finding the formula for the area of a cross-section of $S$ perpendicular to the $x$-axis at a specific $x$.

The integral to find the area is: $A(x) = \int_{y=1}^{y=4} 2k x^2 \sqrt{y} \, dy$

• When $k = 1$, compute: $A(x) = \int_{1}^{4} 2x^2 \sqrt{y} \, dy$ The integration will give the formula for the cross-sectional area.

Similarly, substitute $k = 2$ and repeat.

(b) Cross-sections perpendicular to the $y$-axis

Here, the area of a cross-section perpendicular to the $y$-axis is given by: $A(y) = \int_{x=0}^{x=2} 2x^2 \sqrt{k} \, dx$ You’ll solve this integral to compute cross-sectional areas for different values of $k$.

• For $k = 1$ and $k = 3$, evaluate the integral.

(c) Volume of $S$

Finally, the volume $V$ of the region $S$ is computed by double integration: $V = \int_{x=0}^{x=2} \int_{y=1}^{y=4} 2x^2 \sqrt{y} \, dy \, dx$ You’ll integrate with respect to $y$ first and then with respect to $x$.

Would you like me to calculate these integrals step-by-step for you? Let me know if you have specific parts you'd like to focus on.

Here are 5 related questions for further exploration:

1. How do the bounds of the region $R$ in the $xy$-plane affect the limits of integration?
2. Can you interpret the role of the constant $k$ in modifying the cross-sectional areas?
3. How does the surface equation $z = 2x^2 \sqrt{y}$ influence the shape of the solid $S$?
4. What strategies can simplify double integrals for regions like $R$?
5. How can volume formulas differ if $z = f(x, y)$ is replaced with a different function?

Tip: Always integrate in the inner variable (e.g., $y$) first when solving double integrals, as the outer bounds often depend on the inner results.