consider the solid region s that lies under the surface z = 5x² √y and above the rectangle r = 0, 2 × 1, 4…

Answer:

1. Formula for cross - section area: $A(x)=5x^{2}\int_{1}^{4}\sqrt{y}dy$
2. When (k = 1): $\frac{70}{3}$
3. When (k = 2): $\frac{280}{3}$

Explanation:

Step1: Recall cross - section area formula

The area of a cross - section of the solid (S) perpendicular to the (x) - axis at (x) is given by the integral of the function (z = 5x^{2}\sqrt{y}) with respect to (y) over the interval ([1,4]) since (x) is fixed. So (A(x)=\int_{y = 1}^{y = 4}5x^{2}\sqrt{y}dy=5x^{2}\int_{1}^{4}\sqrt{y}dy).

Step2: Evaluate the integral (\int_{1}^{4}\sqrt{y}dy)

We know that (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) for (n\neq - 1). Here (n=\frac{1}{2}), so (\int_{1}^{4}\sqrt{y}dy=\left[\frac{2}{3}y^{\frac{3}{2}}\right]_{1}^{4}=\frac{2}{3}(4^{\frac{3}{2}}-1^{\frac{3}{2}})=\frac{2}{3}(8 - 1)=\frac{14}{3}).

Step3: Find (A(x))

Since (A(x)=5x^{2}\int_{1}^{4}\sqrt{y}dy) and (\int_{1}^{4}\sqrt{y}dy=\frac{14}{3}), then (A(x)=\frac{70}{3}x^{2}).

Step4: Calculate (A(1))

When (x = k=1), (A(1)=\frac{70}{3}\times1^{2}=\frac{70}{3}).

Step5: Calculate (A(2))

When (x = k = 2), (A(2)=\frac{70}{3}\times2^{2}=\frac{280}{3}).