The region under the curves #y=sqrt(e^x+1), 0<=x<=3# is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

Answer 1

Please see below. (There is only one solid.)

To sketch #y = sqrt(e^x+1)# note that the #y# intercept is #sqrt(2) ~~ 1.4#.

And at #x=3#, we have #y= sqrt(e^3+1)# which is close to #4.5#.

Arithmetic approximation:
(#e~~2.7# so #e^3 ~~ 2.7^3#. now, #2.7 * 2.7 ~~7.3# and #2.7 * 7.3 ~~19.7#. So #e^3 +1# is near #21# and #4.5^2 = 20.25# since #(n+0.5)^2 = n(n+1) +0.25#.)

The derivative: #y' = e^x/(2sqrt(e^x+1))# is always positive, so the function is increasing.

(For a better sketch investigate concavity. The graph is concave up.)

The region is shaded blue.

To go around the #x# axis, I used disks.

The volume of a representative disk is

#pi("radius")^2 xx "thickness" = pi(sqrt(e^x+1))^2 dx = pi(e^x+1) dx#

Since #x# varies from #0# to #3#, the volume of the solid is

#V = pi int_0^3(e^x+1) dx = pi[e^x+x]_0^3 = pi(e^3+2)#

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Answer 2

To sketch the region under the curve ( y = \sqrt{e^x + 1} ) for ( 0 \leq x \leq 3 ) and find the volumes of the two solids of revolution when this region is rotated about the x-axis, follow these steps:

  1. Sketch the curve ( y = \sqrt{e^x + 1} ) for ( 0 \leq x \leq 3 ). This will give you the upper boundary of the region.

  2. Since the lower boundary is the x-axis, the region is bounded by the curve ( y = \sqrt{e^x + 1} ) and the x-axis for ( 0 \leq x \leq 3 ).

  3. To find the volumes of the two solids of revolution, divide the region into two parts at the point where the curve intersects the x-axis. This point can be found by setting ( \sqrt{e^x + 1} = 0 ) and solving for x. However, since the square root of a positive number is always positive, the curve does not intersect the x-axis for ( 0 \leq x \leq 3 ). This means that the entire region is above the x-axis.

  4. The first solid of revolution is generated by rotating the region above the x-axis about the x-axis. To find its volume, use the formula ( V_1 = \pi \int_{0}^{3} [\sqrt{e^x + 1}]^2 , dx ).

  5. The second solid of revolution is the region below the x-axis (which in this case is empty) rotated about the x-axis. Its volume is given by ( V_2 = \pi \int_{0}^{3} [\sqrt{e^x + 1}]^2 , dx ).

  6. Evaluate the integrals to find the volumes of the two solids.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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