this question is designed to be answered without a calculator. the region in the first quadrant bounded by y…

this question is designed to be answered without a calculator. the region in the first quadrant bounded by y = x³, x = 0, and y = 8 is revolved about the x - axis. which definite integral represents the volume of this solid? o π∫₀⁸(8 - x³)²dx o π∫₀⁸(8²-(x³)²)dx o π∫₀²(8 - x³)²dx o π∫₀²(8²-(x³)²)dx

this question is designed to be answered without a calculator. the region in the first quadrant bounded by y = x³, x = 0, and y = 8 is revolved about the x - axis. which definite integral represents the volume of this solid? o π∫₀⁸(8 - x³)²dx o π∫₀⁸(8²-(x³)²)dx o π∫₀²(8 - x³)²dx o π∫₀²(8²-(x³)²)dx

Answer

Explanation:

Step1: Find intersection x - value

Set $y = x^{3}=8$, solve for $x$. We get $x = 2$ since $2^{3}=8$.

Step2: Use disk - washer method

The volume $V$ of the solid of revolution about the $x$ - axis using the disk - washer method is $V=\pi\int_{a}^{b}(R^{2}-r^{2})dx$, where $R$ is the outer radius and $r$ is the inner radius. Here, the outer radius $R = 8$ and the inner radius $r=x^{3}$, and the limits of integration are from $x = 0$ to $x = 2$. So $V=\pi\int_{0}^{2}(8^{2}-(x^{3})^{2})dx$.

Answer:

$\pi\int_{0}^{2}(8^{2}-(x^{3})^{2})dx$