Calculating Volume of Solid Rotated about Y-axis from Bounded Curves

The bound at 4.198 is when (x-3)^2 = lnx, which does not quite happen for these bounds.In summary, the conversation discusses finding the volume of a solid obtained by rotating a region bounded by the curves y= lnx and y=( x-3)^2 about the y-axis. The use of the washer and shell methods are mentioned, with the latter being the recommended approach. The correct setup for the shell method is given and the volume is calculated. Slight discrepancies in the bounds are noted.
  • #1
mxthuy95
3
0

Homework Statement



Consider the region bounded by the curves y= lnx and y=( x-3)^2

Find the volume of the solid obtained by rotating the region about the y-axis


Homework Equations





The Attempt at a Solution



For this I solve for the x so i got x= e^y and x= (y)^(1/2) +3

and for volume equation i got :
1.47
∫ ((y)^(1/2) +3)^2 - (e^y)^2 = 12.481(pi). My answer is
0

different from the book. So can someone point out wat i done wrong and help me to get the right answer

Thank you
 
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  • #2
I cannot see what you did wrong because you did not show your working - you went right from the integral to the final numerical value ... and the integration you show does not look like the one for volume of rotation.

So I cannot tell what you were doing.
I suspect you need to sketch the two curves in the x-y plane and shade in the region bounded by them. The curve ##y=(x-3)^2## intersects the curve ##x=e^y## in two places. ##x=\sqrt{y}+3## only gives you part of the curve you need.
 
Last edited:
  • #3
You appear to be using the 'washer' method, i.e. a stack of elemental discs. This isn't going to work as a single range of integration because there's a sudden change of behaviour at the lower solution of ln(x) = (x-3)2. (Try drawing the region, as Simon advises.) I suggest the cylinder method.
 
  • #4
Hi

Is this the right set up for the shell method

radius = x and height = lnx- (x-3)^2 so i get

A= 2(pi)x((x)(lnx-(x-3)^2) = 2(pi)(xlnx - (x)(x-3)^2 )so volume =

4.170
2(pi) ∫ (xlnx - (x)(x-3)^2
2.127

Thanks
 
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  • #5
mxthuy95 said:
4.170
2(pi) ∫ (xlnx - (x)(x-3)^2
2.127
Yes, except that I get slightly different bounds: 2.130, 4.198.
 

1. What is the formula for calculating the volume of a solid rotated about the y-axis from bounded curves?

The formula for calculating the volume of a solid rotated about the y-axis from bounded curves is V = π∫[a,b] (f(x))² dx, where a and b are the limits of integration and f(x) is the function defining the bounded curves.

2. Can this formula be used to calculate the volume of any solid rotated about the y-axis?

Yes, this formula can be used to calculate the volume of any solid rotated about the y-axis as long as the solid is bounded by curves that can be represented by a function f(x).

3. How do I determine the limits of integration for the formula?

The limits of integration can be determined by finding the x-values where the bounded curves intersect. These x-values will be the limits of integration for the integral.

4. Are there any restrictions on the type of curves that can be used in this formula?

The curves used in this formula must be continuous and have a defined function f(x). Additionally, the curves must be bounded, meaning they have a beginning and an end, and they cannot intersect or overlap each other.

5. Is there a specific unit that should be used for the volume calculation?

The unit for the volume calculation can vary depending on the units used for the bounded curves. However, it is important to ensure that all units are consistent throughout the calculation. If the bounded curves are in meters, for example, the volume should be in cubic meters.

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