Question

Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate​ axes, the curve y equals e Superscript x​, and the line x equals ln 16 about the line x equals ln 16.

Answer 1

Answer:

$\large \boxed{2\pi(15 - \ln16)}$

Step-by-step explanation:

A 90° portion of your volume is shown below.

Let's use the shell method and decompose the solid of revolution into cylindrical shells.

The volume of each cylindrical shell is

dV = 2πrh dx

where  

r = ln16 - x

h = eˣ

So

dV = 2π(ln16 - x)eˣdx  

$\begin{array}{rcl}V& =& 2\pi \int_{0}^{\ln16}(\ln16 - x)e^{x}dx\\\\&=&2\pi \int_{0}^{\ln16}(\ln 16)e^{x}dx -2\pi \int_{0}^{\ln16} xe^{x}dx\\\\&=& 2\pi \ln16 \int_{0}^{\ln16}e^{x}dx - 2\pi\int_{0}^{\ln16}xe^{x}dx\\\\\end{array}$

Let's integrate the two terms separately.

$\begin{array}{rcl}A &=& 2\pi \ln16 \int_{0}^{\ln16}e^{x}dx\\\\&=& 2\pi\ln16[e^{x}]_{0}^{\ln16}\\\\&=& 2\pi\ln16(16 - 1)\\&= &\mathbf{30\pi\ln1}\\\\B &=&2\pi\int_{0}^{\ln16}xe^{x}dx\\\end{array}$

Use the integration by parts formula

$\int u dv = uv - \int v du$

$\begin{array}{rcl}uv - \int v du &=&\\B &=& 2\pi[xe^{x}]_{0}^{\ln16} - 2\pi \int_{0}^{\ln16} e^{x}dx\\\\&=&2\pi[xe^{x}]_{0}^{\ln16}-2\pi [e^{x}]_{0 }^{\ln16}\\\\&= &2\pi(16\ln16- 0) -2\pi(16 - 1)\\&=& \mathbf{32\pi \ln16 -30\pi}\\A - B &= &30\pi\ln16 - 32\pi \ln16 +30\pi\\&=& 30\pi - 2\pi \ln16\\&=&\mathbf{2\pi(15 - \ln16)}\end{array}\\\text{The volume of the solid is \large \boxed{\mathbf{2\pi(15 - \ln16)}} }$