what is the volume of the solid generated when the region bounded by the graph of $y = x^{3}$, the vertical…

what is the volume of the solid generated when the region bounded by the graph of $y = x^{3}$, the vertical line $x = 4$, and the horizontal line $y = 8$ is revolved about the horizontal line $y = 8$?\n\na $piint_{2}^{4}(x^{3}-8)dx$\n\nb $piint_{2}^{4}(x^{6}-64)dx$\n\nc $piint_{2}^{4}(x^{3}-8)^{2}dx$

what is the volume of the solid generated when the region bounded by the graph of $y = x^{3}$, the vertical line $x = 4$, and the horizontal line $y = 8$ is revolved about the horizontal line $y = 8$?\n\na $piint_{2}^{4}(x^{3}-8)dx$\n\nb $piint_{2}^{4}(x^{6}-64)dx$\n\nc $piint_{2}^{4}(x^{3}-8)^{2}dx$

Answer

Explanation:

Step1: Find the intersection point

Set $y = x^{3}=8$, then $x = 2$.

Step2: Use the disk - washer method

When a region bounded by $y = f(x)$ and $y = c$ is revolved about the line $y = c$, the volume $V$ of the solid of revolution using the disk - washer method is given by $V=\pi\int_{a}^{b}[f(x)-c]^{2}dx$. Here, $f(x)=x^{3}$, $c = 8$, $a = 2$ and $b = 4$. So the volume $V=\pi\int_{2}^{4}(x^{3}-8)^{2}dx$.

Answer:

C. $\pi\int_{2}^{4}(x^{3}-8)^{2}dx$