Volume bounded by graphs

In summary: Here is a picture of the presumed region, assuming the upper boundary is ##y = \frac \pi 2##. The actual region continues on infinitely far to the right as the arctan nears its asymptote.LCkurtz that is what the question said exactly straight to the teeth from my course pack.Once again word for word. Set up the integral (BUT DO NOT SOLVE) for the volume of the object created by rotating the region bounded by y = arctan(x)and y = arcsin(x) in the first quadrant.a.) rotate about the x axisb.) rotate about the y axisThat is word for work LCkurtz. You see I chose my boundary
  • #1
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Homework Statement



Set up the integral (but do not solve) for the volume of the object created by rotating the region bounded by y = arctan(x) and y = arcsin(x) in the first quadrant.

Homework Equations



I = ∏∫(f(x)^2 - g(x)^2) dx

The Attempt at a Solution




a.) rotate about he x axis

I came up with

I = ∏∫(arcsin(x))^2 - (arctan(x))^2 dx between [0,1]

b.) rotate about the y -axis

I came up with

I = ∏∫(arctan(x))^2 - (arcsin(x))^2 dx between [0,∏/2]

Did I do this correctly.
Thank you.
 
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  • #2
Lets start with part a).

Re-write your functions as ##x(y)##, otherwise you won't be able to find the limits of integration ( The only solution to ##arcsin(x) = arctan(x)## is ##x=0##. ##x=1## is not a solution. ).

The solid formed from this process will be sort of diamond shaped.
 
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  • #3
Jbreezy said:

Homework Statement



Set up the integral (but do not solve) for the volume of the object created by rotating the region bounded by y = arctan(x) and y = arcsin(x) in the first quadrant.

Those two curves do not enclose an area in the first quadrant.
 
  • #4
LCKurtz said:
Those two curves do not enclose an area in the first quadrant.

LCkurtz that is what the question said exactly straight to the teeth from my course pack.


Once again word for word.

Set up the integral (BUT DO NOT SOLVE) for the volume of the object created by rotating the region bounded by y = arctan(x)and y = arcsin(x) in the first quadrant.
a.) rotate about the x axis
b.) rotate about the y axis

That is word for work LCkurtz. You see I chose my boundary for revolving around x to be 1 because the domain of arcsin(x) goes form -1 to 1. And I chose ∏/2 when revolving around the y-axis because the range of arcsin(x) and arctan(x).
Those are my thoughts.
 
  • #5
Jbreezy said:
LCkurtz that is what the question said exactly straight to the teeth from my course pack.Once again word for word.

Set up the integral (BUT DO NOT SOLVE) for the volume of the object created by rotating the region bounded by y = arctan(x)and y = arcsin(x) in the first quadrant.
a.) rotate about the x axis
b.) rotate about the y axis

That is word for work LCkurtz. You see I chose my boundary for revolving around x to be 1 because the domain of arcsin(x) goes form -1 to 1. And I chose ∏/2 when revolving around the y-axis because the range of arcsin(x) and arctan(x).
Those are my thoughts.

That's why I said, you have to change the functions into functions of ##y## and use a different method entirely. You won't be able to solve this with the usual method you've been using. The functions as is in the problem only have ONE intersection and therefore do not bound any particular area.
 
  • #6
Jbreezy said:
LCkurtz that is what the question said exactly straight to the teeth from my course pack.


Once again word for word.

Set up the integral (BUT DO NOT SOLVE) for the volume of the object created by rotating the region bounded by y = arctan(x)and y = arcsin(x) in the first quadrant.
a.) rotate about the x axis
b.) rotate about the y axis

That is word for work LCkurtz. You see I chose my boundary for revolving around x to be 1 because the domain of arcsin(x) goes form -1 to 1. And I chose ∏/2 when revolving around the y-axis because the range of arcsin(x) and arctan(x).
Those are my thoughts.

I don't care if it is "straight to the teeth", written in gold ink, and blessed by the Pope. Those two curves do not enclose a region in the first quadrant. A reasonable GUESS might be to add the line ##y = \frac \pi 2## to the area description. That would define a region that is infinite in extent requiring improper integrals. You have to properly define your region before you try setting up any integrals.
 
  • #7
Haha. The Pope doesn't belong in physics forum come on!
But seriously though I said the upper bound for the integral was PI/2. But because the PI/2 is not included in the range of arctan(x) does this mean that it is infinite in extent? What about the the upper limit of integration I gave when it is around the x axis? I said I wanted it to go from 0 to 1 because that is the range of arcsin(x).
I'm kind of lost with what to do because like I said this is straight from my course pack.
 
  • #8
Jbreezy said:
Haha. The Pope doesn't belong in physics forum come on!
But seriously though I said the upper bound for the integral was PI/2. But because the PI/2 is not included in the range of arctan(x) does this mean that it is infinite in extent? What about the the upper limit of integration I gave when it is around the x axis? I said I wanted it to go from 0 to 1 because that is the range of arcsin(x).
I'm kind of lost with what to do because like I said this is straight from my course pack.

Here's a graph to help you understand why you can't do this the way you're thinking :

http://gyazo.com/f163b065b07aa3a19ab25798611de7c6

The graphs will never intersect again as ##x → ∞##. Therefore there is no other point of intersection as to form a closed area with.

Re-writing you get ##x = sin(y)## and ##x = tan(y)##.

When does ##sin(y) = tan(y)##? One point is ##y=0##. Solving the rest will give you your limits for ##y##. Then you need to integrate with respect to ##y## and not ##x##.
 
  • #9
OK, but I have one method to use which is the disk method. If I use the disk method and revolve it around the x-axis wouldn't I need to integrate with respect to x?
 
  • #10
Here is a picture of the presumed region, assuming the upper boundary is ##y = \frac \pi 2##. The actual region continues on infinitely far to the right as the arctan nears its asymptote.

shaded.jpg
 
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  • #11
Thanks, I see what you are saying but how am I supposed to do this problem then? Revolving it around the x-axis and using the disk method? Maybe I could do it with respect to x and go from 0 to 1 then go from 1 to infinity? I mean this makes no sense as a question considering the chapter of information I'm supposed to use.
 
  • #12
Jbreezy said:
Thanks, I see what you are saying but how am I supposed to do this problem then? Revolving it around the x-axis and using the disk method? Maybe I could do it with respect to x and go from 0 to 1 then go from 1 to infinity? I mean this makes no sense as a question considering the chapter of information I'm supposed to use.

Yes, if you want to use the disk method for revolving about the x axis, you would need two integrals just as you say from 0 to 1 and from 1 to infinity, because the upper radius changes.

You can do the disk method around the y-axis with one integral. Show us what you get for each.
 
  • #13
Well I'm just supposed to set up the integral not actually do it.

This is what I set up though.For about x I have
I_1 = ∏∫(arcsin(x))^2 - (arctan(x))^2 dx from 0 to 1.
I_2 = ∏∫((∏/2)^2 - (arctan(x))^2 dx and from 1 to infinity

Volume is then I_1 + I_2 = A

About y

I = ∏∫(tan(x))^2 - (sin(x))^2 dy from 0 to ∏/2.

EDIT: I think it is OK now.
 
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  • #14
Jbreezy said:
Well I'm just supposed to set up the integral not actually do it.

This is what I set up though.


For about x I have
I_1 = ∏∫(arcsin(x))^2 - (arctan(x))^2 dx from 0 to 1.
I_2 = ∏∫((∏/2)^2 - (arctan(x))^2 dx and from 1 to infinity

Area is then I_1 + I_2 = A

You mean volume, not area.

About y

I = ∏∫(arctan(x))^2 - (arcsin(x))^2 dx from 0 to ∏/2.

Your one about x is OK, but the second one isn't. You need things in terms of y and dy integrals.
 
  • #15
OK look at the post again. I did mean volume and yes I forgot to put it in terms of y.
 
  • #16
Jbreezy said:
OK look at the post again. I did mean volume and yes I forgot to put it in terms of y.

No, close but still not right. It should be of the form ##\int x_{right}^2 - x_{left}^2~dy## where the ##x## values are given ##x## in terms of ##y## from the curves.
 
  • #17
About y

I = ∏∫(tan(y))^2 - (sin(y))^2 dy from 0 to ∏/2.

? Opps
 
  • #18
good?
 
  • #19
Jbreezy said:
About y

I = ∏∫(tan(y))^2 - (sin(y))^2 dy from 0 to ∏/2.

?

That's correct now.
 
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  • #20
Yeah I just totally didn't pay attention my brain said one thing hand did the other.
Thanks
 

1. What is the definition of "volume bounded by graphs"?

The volume bounded by graphs refers to the space enclosed between two or more intersecting graphs on a coordinate plane. It can also be thought of as the three-dimensional region under a surface or between two surfaces.

2. How is the volume bounded by graphs calculated?

The volume bounded by graphs can be calculated using integration techniques from calculus. For a region bounded by two graphs, the integral would be taken from the lower bound of the region to the upper bound, with the appropriate functions for the two graphs being used in the integrand.

3. What are some real-life applications of finding volume bounded by graphs?

Finding volume bounded by graphs has many practical applications, such as calculating the volume of a swimming pool, determining the capacity of a storage container, or finding the volume of a chemical reaction in a beaker. It is also used in engineering and architecture for designing structures and determining the amount of material needed for construction.

4. Can the volume bounded by graphs be negative?

No, the volume bounded by graphs cannot be negative. Volume is a measure of space, and it is always a positive value. In some cases, the bounded volume may equal zero if the two graphs do not intersect or if they are identical.

5. Are there any limitations to calculating volume bounded by graphs?

One limitation is that the graphs must be continuous and smooth, otherwise, the calculated volume may not be accurate. Additionally, the method of integration used may be limited by the complexity of the graphs, which can make the calculation more challenging.

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