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 the y-axis. round your answer to three decimal places.

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.

college