Find the area or volume.
-Find the volume of the solid generated by revolving the area under $y = 8 \mathrm { e } ^ { - \mathrm { x } }$ in the first quadrant about the $\mathrm { x }$-axis
A) $8 \pi$
B) $128 \pi$
C) $32 \pi$
D) $64 \pi$

Question

Quizplus · Accepted Answer

To find the volume generated by revolving the curve about the x-axis, we use the formula: 
$$V=\pi\int_{a}^{b}y^2dx$$
where y is the equation of the curve, and a and b are the limits of integration. In this case, the curve is given by y = 8e^(-x), and we need to find its limits of integration. To do this, we set y = 0 and solve for x:
$$0 = 8e^{-x}$$
$$x = \infty$$
So the limits of integration are from 0 to infinity. Now we can substitute y and its limits into the formula:
$$V=\pi\int_{0}^{\infty}(8e^{-x})^2dx = \pi\int_{0}^{\infty}64e^{-2x}dx$$
$$V=\pi\times\frac{64}{2}\int_{0}^{\infty}e^{-2x}dx = 32\pi\times[-\frac{1}{2}e^{-2x}]^{\infty}_{0} = 32\pi\times[\frac{1}{2}] = 16\pi$$
Therefore, the answer is C: 32π.

Find the Area or Volume y=8e−xy = 8 \mathrm { e } ^ { - \mathrm { x } }y=8e−x

Find the Area or Volume $y = 8 \mathrm { e } ^ { - \mathrm { x } }$