To find the volume of the solid generated by revolving the region enclosed by the graph of e^(x/2), y=1, and x=ln(3) around the x-axis, we can indeed use either the washer method or the disk method.
The washer method is typically used when we have a region bounded by two curves, while the disk method is used when we have a region bounded by one curve and two lines. In this case, we have the graph of e^(x/2) on one side and the line y=1 on the other side, so both methods are applicable.
Let's use the disk method to find the volume of the solid generated:
1. First, let's find the limits of integration. We are revolving the region around the x-axis, so we need to find the x-values where the graph of e^(x/2) and y=1 intersect.
Setting e^(x/2) equal to 1, we have:
e^(x/2) = 1
Taking the natural logarithm of both sides, ln(e^(x/2)) = ln(1)
x/2 = 0
x = 0
Setting y=1 equal to 1, we have:
1 = 1
This tells us that the region starts and ends at x=0 and x=ln(3).
2. Next, let's find the radius of each disk. Since we are revolving the region around the x-axis, the radius of each disk is simply the height of the curve at each x-value. In this case, the height is given by e^(x/2).
3. Now, let's set up the integral for the volume using the disk method:
V = ∫[a,b]π(r(x))^2 dx
where [a,b] represents the limits of integration (in this case, from x=0 to x=ln(3)), r(x) represents the radius of each disk, and dx represents an infinitely small change in x.
So the integral becomes:
V = ∫[0,ln(3)]π(e^(x/2))^2 dx
V = ∫[0,ln(3)]πe^x dx
4. Finally, evaluate the integral to find the volume:
V = π∫[0,ln(3)]e^x dx
To evaluate this integral, we can use the power rule for integration, which states that ∫e^x dx = e^x + C.
V = π(e^x)|[0,ln(3)]
V = π(e^ln(3) - e^0)
Simplifying further:
V = π(3 - 1)
V = 2π
Therefore, the volume of the solid generated by revolving the region enclosed by the graph of e^(x/2), y=1, and x=ln(3) around the x-axis is 2π cubic units.