Let R be the region in the first quadrant enclosed by the lines #x=ln 3# and #y=1# and the graph of #y=e^(x/2)#, how do you find the volume of the solid generated when R is revolved about the line y=-1?

Answer 1

# V = pi( -2 + 4 sqrt 3 - 3 ln 3 ) #

Easiest is to revolve #e^{x/2}# about y = -1 and then deduct the cylinder (shaded green) which has volume #pi * 2^2 * ln 3 = 4 pi ln 3#

So the volume of the small strip width dx when #e^{x/2}# is revolved about y = -1 is

#dV = pi (e^{x/2} + 1)^2 dx#

#V = pi int_0^{ln 3} dx qquad (e^{x/2} + 1)^2 #

# = pi int_0^{ln 3} dx qquad e^{x} + 2 e^{x/2} + 1 #

# pi [e^{x} + 4 e^{x/2} + x ]_0^{ln 3}#

# pi( [3 + 4 sqrt 3 + ln 3 ] - [1 + 4 + 0] )#

# pi( -2 + 4 sqrt 3 + ln 3 ) #

Then deducting the cylinder (shaded green) which has volume #= 4 pi ln 3#:

# V = pi( -2 + 4 sqrt 3 - 3 ln 3 ) #

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Answer 2

To find the volume of the solid generated when the region ( R ) is revolved about the line ( y = -1 ), you can use the method of cylindrical shells. The volume ( V ) is given by the integral:

[ V = 2\pi \int_{a}^{b} r(x) \cdot h(x) , dx ]

Where:

  • ( r(x) ) represents the distance from the axis of revolution to the shell at position ( x ),
  • ( h(x) ) represents the height of the shell at position ( x ),
  • ( a ) and ( b ) are the limits of integration.

In this case, the region ( R ) is bounded by the lines ( x = \ln 3 ), ( y = 1 ), and the graph ( y = e^{x/2} ). The limits of integration ( a ) and ( b ) will be the x-coordinates of the intersection points between the curve ( y = e^{x/2} ) and the line ( y = 1 ).

To set up the integral, you'll need to express ( r(x) ) and ( h(x) ) in terms of ( x ). ( r(x) ) is the distance from ( y = -1 ) to the curve ( y = e^{x/2} ), which is ( 1 + e^{x/2} ). ( h(x) ) is the length of the cylindrical shell, which is ( \ln 3 - x ).

So, the integral becomes:

[ V = 2\pi \int_{\ln 3}^{b} (1 + e^{x/2})(\ln 3 - x) , dx ]

You then need to find the value of ( b ) by solving ( e^{x/2} = 1 ), which yields ( x = 0 ). Thus, the integral becomes:

[ V = 2\pi \int_{\ln 3}^{0} (1 + e^{x/2})(\ln 3 - x) , dx ]

You can then evaluate this integral to find the volume of the solid.

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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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