Solve the problem.
-Find the volume of the solid generated by revolving the region in the first quadrant bounded by $y = e ^ { x }$ and the $x$-axis, from $x = 0$ to $x = \ln 3$, about the $y$-axis.
A) $6 \pi \ln 3$
B) $2 \pi ( 3 \ln 3 - 2 )$
C) $2 \pi ( 3 \ln 3 - 3 )$
D) $3 \pi \ln 3$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region bounded by $y = e^x$ and the $x$-axis, from $x = 0$ to $x = \ln 3$, about the $y$-axis can be found using the method of cylindrical shells. The formula for the volume of a solid of revolution generated by revolving a function $y = f(x)$ about the $y$-axis from $x = a$ to $x = b$ is given by $V = 2\pi \int_{a}^{b} x f(x) dx$. Substituting $f(x) = e^x$, $a = 0$, and $b = \ln 3$, we get:$$V = 2\pi \int_{0}^{\ln 3} x e^x dx$$To solve this integral, we use integration by parts, where $u = x$ and $dv = e^x dx$, which gives $du = dx$ and $v = e^x$. Thus, the integral becomes:$$V = 2\pi \left[ x e^x \bigg|_{0}^{\ln 3} - \int_{0}^{\ln 3} e^x dx \right]$$$$V = 2\pi \left[ (\ln 3)e^{\ln 3} - 0 - (e^{\ln 3} - e^0) \right]$$$$V = 2\pi \left[ 3\ln 3 - (3 - 1) \right]$$$$V = 2\pi (3\ln 3 - 2)$$Therefore, the correct answer is $2\pi (3\ln 3 - 2)$, which is choice B.

Solve the Problem y=exy = e ^ { x }y=ex And The

Solve the Problem $y = e ^ { x }$ And The