(b) find a formula for the area of a cross - ...

# Answer: (b) Let's assume that the cross - section perpendicular to the $y$-axis at $y = k$ is a rectangle. If we need to find the area of the cross - section $A(k)$ of the solid $S$ in the plane perpendicular to the $y$-axis at $y=k$ ($1\leq k\leq4$). Suppose the base of the rectangle in the $x - z$ plane (for a fixed $y = k$) has length in the $x$ - direction from $x = 0$ to $x = 2$ and the height in the $z$ - direction is some function of $y$. Without more information about the height of the cross - section as a function of $y$, if we assume the height of the cross - section $h(k)$ is a linear function of $y$ (a simple case), and if the height at $y = 1$ is $z_1$ and at $y = 4$ is $z_2$. First, we find the linear relationship $h(k)=m(k - 1)+z_1$, where $m=\frac{z_2 - z_1}{3}$. The area of the cross - section $A(k)$ (a rectangle with base $b = 2$ and height $h(k)$) is $A(k)=2h(k)$. If we assume the height of the cross - section at $y = 1$ is $1$ and at $y = 4$ is $3$, then $m=\frac{3 - 1}{3}=\frac{2}{3}$, and $h(k)=\frac{2}{3}(k - 1)+1=\frac{2k-2 + 3}{3}=\frac{2k + 1}{3}$. So $A(k)=\frac{4k + 2}{3}$. When $k = 1$: $A(1)=\frac{4\times1+2}{3}=2$ When $k = 3$: $A(3)=\frac{4\times3+2}{3}=\frac{12 + 2}{3}=\frac{14}{3}$ (c) The volume $V$ of the solid $S$ using the cross - sectional area formula $V=\int_{a}^{b}A(y)dy$, where $a = 1$, $b = 4$ and $A(y)=\frac{4y + 2}{3}$. \[ \begin{align*} V&=\int_{1}^{4}\frac{4y + 2}{3}dy\\ &=\frac{1}{3}\int_{1}^{4}(4y+2)dy\\ &=\frac{1}{3}\left[4\times\frac{y^{2}}{2}+2y\right]_{1}^{4}\\ &=\frac{1}{3}\left[(2y^{2}+2y)\big|_{1}^{4}\right]\\ &=\frac{1}{3}\left[(2\times4^{2}+2\times4)-(2\times1^{2}+2\times1)\right]\\ &=\frac{1}{3}\left[(32 + 8)-(2 + 2)\right]\\ &=\frac{1}{3}(40 - 4)\\ &=12 \end{align*} \] # Explanation: ## Step1: Assume cross - section shape Assume cross - section is rectangle. ## Step2: Find height function Find linear height function $h(k)$. ## Step3: Calculate cross - sectional area $A(k)=2h(k)$. ## Step4: Evaluate cross - sectional areas Substitute $k = 1$ and $k = 3$ into $A(k)$. ## Step5: Set up volume integral $V=\int_{1}^{4}A(y)dy$. ## Step6: Integrate Integrate $\frac{1}{3}\int_{1}^{4}(4y + 2)dy$. ## Step7: Evaluate definite integral Find value of $V$.