find the volume of the solid generated by revolving around the x-axis: y=e^(x-8), y=0, x=0, x=10
So I know you can use the washer method, but the shell method can also be used and it should be better. The issue is I'm not certain how to actually do it. Any help is appreciated!
washers:
v = ∫[0,10] πr^2 dx
where r = y = e^(x-8)
v = π∫[0,10] (e^(x-8))^2 dx
= π∫[0,10] e^(2x-16) dx
= π/2 e^(2x-16) [0,10]
= π/2 (e^4 - 1/e^16)
shells:
y = e^(x-8), so x = 8+lny
v = ∫[e^-8,e^2] 2πrh dy
where r = y and h = 10-x = 2-lny
v = 2π∫[e^-8,e^2] y(2-lny) dy
= π/2 (y^2(5-2lny)) [e^-8,e^2]
= π/2 (e^4 - 21/e^16)
Hmmm. better double-check my math. Anyway, that's the method.
To find the volume of the solid generated by revolving the curve y = e^(x-8) around the x-axis, you can indeed use either the washer method or the shell method.
Since you mentioned that the shell method might be better suited for this particular problem, let's go ahead and use that method.
The shell method involves integrating the circumferences of cylindrical shells along the interval where the solid is being revolved and summing up these infinitesimally thin shells to find the total volume.
Here's how you can apply the shell method step by step to find the volume:
1. Determine the limits of integration:
The curve y = e^(x-8) intersects the x-axis at y = 0, so these are the lower and upper limits of integration. From the given information, the interval is x = 0 to x = 10.
2. Set up the integral:
The formula for the volume using the shell method is V = 2π * integral[R(x) * H(x) * dx], where R(x) is the radius of the shell and H(x) is the height or length of the shell.
For this problem, R(x) will be the distance from the x-axis to the curve, and H(x) will be an infinitesimally small change in x.
3. Express R(x) and H(x) in terms of x and evaluate the integral:
The distance from the x-axis to the curve is simply the function value at each x. Therefore, R(x) = e^(x-8), and since H(x) is an infinitesimally small change in x, H(x) can be represented as dx.
The integral then becomes:
V = 2π * integral[e^(x-8) * dx] from x = 0 to x = 10.
4. Solve the integral:
Integrating e^(x-8) with respect to x gives us:
V = 2π * [e^(x-8)] evaluated from x = 0 to x = 10.
To evaluate this, plug in the upper limit of 10 and subtract the value of the function at the lower limit of 0:
V = 2π * [e^(10-8) - e^(0-8)].
Simplifying, we get:
V = 2π * [e^2 - e^(-8)].
And that's the volume of the solid generated by revolving y = e^(x-8) around the x-axis using the shell method.