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.

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -