How do you find the volume of the solid with base region bounded by the curve #y=e^x#, #y=ln4#, and the #y#-axis if cross sections perpendicular to the #y#-axis are squares?

Answer 1

Since its cross-sections are squares, its area can be expressed as
#A(y)=(lny)^2#
So, the volume of the solid can be found by:
#V=int_1^{ln4}(lny)^2dy#.
Can you go from here? You might have to use integration by parts twice.
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Answer 2

To find the volume of the solid with base region bounded by the curves ( y = e^x ), ( y = \ln 4 ), and the y-axis, where the cross sections perpendicular to the y-axis are squares, integrate the area of each square cross section along the y-axis.

The width of each square cross section is the difference between the upper and lower y-values of the region, which is ( e^x - \ln 4 ). The area of each square is then ( (e^x - \ln 4)^2 ).

Integrate this area with respect to ( y ) over the interval from ( \ln 4 ) to ( e^x ), which gives the volume of the solid:

[ V = \int_{\ln 4}^{e^x} (e^x - \ln 4)^2 , dy ]

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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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