Answer:
[tex]\large \boxed{2\pi(15 - \ln16)}[/tex]
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 = 2Ï€rh dx
where Â
r = ln16 - x
h = eˣ
So
dV = 2Ï€(ln16 - x)eË£dx Â
[tex]\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}[/tex]
Let's integrate the two terms separately.
[tex]\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}[/tex]
Use the integration by parts formula
[tex]\int u dv = uv - \int v du[/tex]
[tex]\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 $\large \boxed{\mathbf{2\pi(15 - \ln16)}}$}[/tex]
