Consider the closed region bounded by $y = 5e^{−x}$, $y = 5$, and $ x= 4$.
In each case, write an integral giving the volume of the solid.
(a)rotate the region around the x-axis.
Can anyone give me some help on this problem? The limits of integration are from 0 to 4 but I am having trouble determining the radius. To me, it seems that the radius is $5e^{-x}$ but that is incorrect. Can someone lead me in the right direction to finding the correct radius?
On
The upper bound of the integral should be $5$ because the area is bounded by $y = 5$.
The lower bound of the integral should be $5e^{-4}$ because the area is bounded by $x = 4$.
The height of our shell should be $4 - (-ln(\frac{y}{5}))$ (you can find it out if you plot the graph of $y = 5e^{-x}$ and find the bounded area.)
Shell Method: $$\int_{5e^{-4}}^{5}2\pi y(4+ln(\frac{y}{5}))dy$$
If you take a cross-section perpendicular to the $x$-axis, you will get a ring with an outer radius ($r_1=5$) and an inner radius ($r_2=5e^{-x}$). The area of that ring is $$A(x) = \pi r_1^2 - \pi r_2^2 = \pi 5^2 - \pi (5e^{-x})^2.$$
Can you take it from here?