consider the solid region s that lies under t...
consider the solid region s that lies under the surface z = 5x² √y and above the rectangle r = 0, 2 × 1, 4. (a) find a formula for the area of a cross - section of s in the plane perpendicular to the x - axis at x for 0 ≤ x ≤ 2. (use the value k for x.) use the formula to compute the areas of the cross - sections illustrated. k = 1 70/3 k = 2 280/3
Answer
# 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 \(x = k = 2\), \(A(2)=\frac{70}{3}\times2^{2}=\frac{280}{3}\).