Find the volume of the solid obtained by rotating the region bounded by the x-axis, the y-axis, the line y=2, and y=e^x about th
e y-axis. Round your answer to three decimal places.
2 answers:
Answer:
0.592
Step-by-step explanation:
Graph the region and find the intersections.
x=0 and y=e^x intersect at (0, 1)
x=0 and y=2 intersect at (0, 2)
y=2 and y=e^x intersect at (ln2, 2)
When we revolve the region around the y-axis, we get a solid cone-like shape.
Attempt 1: Let's try using disk method to find the volume. Divide the volume into a stack of thin disks. Each disk has a position y, radius x, and thickness dy. So the volume of each disk is:
dV = π x² dy
The total volume is:
V = ∫ dV
V = ∫ π x² dy
We are integrating with respect to y, so the limits need to be in terms of y. y is between 1 and 2, so:
V = ∫₁² π x² dy
Next, we need to write x in terms of y.
V = ∫₁² π (ln y)² dy
Hmm, this is not an easy integral. In situations like this, we can try shell method instead of disk method.
Attempt 2: In shell method, we divide the volume into a concentric hollow cylinders (kind of like Russian nesting dolls). Each cylinder has a height h, a radius x, and a thickness dx. So the volume of each cylinder is:
dV = 2π x h dx
The height is h = 2 − y. Substituting:
dV = 2π x (2 − y) dx
So the total volume is:
V = ∫ dV
V = ∫ 2π x (2 − y) dx
We are integrating with respect to x, so the limits need to be in terms of x. x is between 0 and ln 2, so:
V = ∫₀ˡⁿ² 2π x (2 − y) dx
Writing y in terms of x:
V = ∫₀ˡⁿ² 2π x (2 − e^x) dx
Simplifying:
V = 2π ∫₀ˡⁿ² (2x − xe^x) dx
V = 2π [ ∫₀ˡⁿ² (2x dx) − ∫₀ˡⁿ² (xe^x dx) ]
The second integral requires integration by parts, but at least it's easier than the last attempt.
∫ u dv = uv − ∫ v du
If u = x and dv = e^x dx, then du = dx and v = e^x.
∫xe^x dx = xe^x − ∫ e^x dx
∫xe^x dx = xe^x − e^x
∫xe^x dx = (x − 1) e^x
Plugging in:
V = 2π [ ∫₀ˡⁿ² (2x dx) − ((x − 1) e^x) |₀ˡⁿ² ]
V = 2π [ x² |₀ˡⁿ² − ((x − 1) e^x) |₀ˡⁿ² ]
V = 2π [ ((ln2)² − 0²) − ((ln2 − 1) e^(ln2) − (0 − 1) e^0) ]
V = 2π [ (ln2)² − (2 (ln2 − 1) + 1) ]
V = 2π [ (ln2)² − (2 ln2 − 2 + 1) ]
V = 2π [ (ln2)² − (2 ln2 − 1) ]
V = 2π [ (ln2)² − 2 ln2 + 1 ]
V ≈ 0.592
We can check our answer by comparing it to the volume of a cone with height 1 and radius ln 2. It should be slightly smaller than the volume we found.
V = 1/3 π r² h
V = 1/3 π (ln2)² (1)
V ≈ 0.503
So our answer makes sense.
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