Help with integration

  • Thread starter CaityAnn
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  • #1
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Homework Statement



Evaluate the integral 4cos(x^2)/ (2((z)^(1/2))) dxdydz, limits for x 2y to 2, for y, 0 to 1 and for z 01, by changing the order of integration.

Homework Equations



Would I need to change the limits in order to accurately change the order of the integration? Im thinking not, I could just switch the dx and dy and Integrate. I guess thats what she is asking.

The Attempt at a Solution



My problem? I am having a heck of a time integrating this function. Please help me with getting started on integrating it.
 

Answers and Replies

  • #2
cristo
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Yes, you will need to change the limits, since the limits for x depend on y. Try drawing a diagram of the region that you are integrating over (a sketch of the region in the x-y plane will do, since z is constant). If you want to swap the order of integration from dxdy to dydx, then you need limits for y which are independent of x. This diagram should help you deduce such limits.
 
  • #3
mjsd
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you sure it is [tex]\cos(x^2)[/tex] and not [tex]\cos^2 x[/tex]?

if it is indeed [tex]\cos(x^2)[/tex] you get a Fresnel integral in x
 
Last edited:
  • #4
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I am sure it is is x^2 in the parenthesis.

So when I solved for z I got z=(2cos(x^2))^(1/2).
Thats right, right?
I graph this and get a cos function parralel to the y axis.
Since my dx limits are 2y and 2, i can graph 2y and 2 on a 2 D graph and see how it would be on the 3D graph. my Dy limits are from 0 to 1.
So if I have to change my limits instead of just rearranging the function like someone here said I should have to do,.. then I will have to solve the x=2y and x=2 for y? giving dy=1/2x and y=2????????
 
  • #5
mjsd
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firstly let me get this correct are you trying to evaluate
[tex]\displaystyle{\int_{0}^{1}\int_{0}^{1}\int_{2y}^{2} \frac{2\cos (x^2)}{\sqrt{z}}\;dx dy dz}[/tex]
 
  • #6
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yes! The problem is, Evlaute that integral by changing the order of integration. Im wondering if I should just switch the dy , dy limits and integrate or if I do that will i need to change my limits. AND, on top of that, Im doubting my integration skills on this problem. *sigh* thanks for you help.
 
  • #7
mjsd
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if you go direct in doing the x integral... you need to know what are Frensel integrals, but of course changing the order of integration will simplify things in this case... follow cristo's advice and sketch the region of integration (ignore z for the moment, do that last and just look at the x and y integrals)
 
  • #8
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yeah... i know.... if you look at my reply to cristo, I have a question about those x and y limits.
Someone PLEASE
 
  • #9
mjsd
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I am sure it is is x^2 in the parenthesis.

So when I solved for z I got z=(2cos(x^2))^(1/2).
Thats right, right?
I graph this and get a cos function parralel to the y axis.
Since my dx limits are 2y and 2, i can graph 2y and 2 on a 2 D graph and see how it would be on the 3D graph. my Dy limits are from 0 to 1.
So if I have to change my limits instead of just rearranging the function like someone here said I should have to do,.. then I will have to solve the x=2y and x=2 for y? giving dy=1/2x and y=2????????
judging by this response, it appears that you don't know what cristo is talking about.....you change the integration limits by first working out what is the actual region R that you are integrating over... THEN using that knowledge to re-write the double integral where you can do the y bit first followed by the x part. so far, you have not even touched any part of the integrand.

for example:
[tex]\int_{0}^{\pi/2} \int_{0}^{x} f(x,y) \;dy\,dx[/tex] is equivalent to
[tex]\int_{0}^{\pi/2} \int_{y}^{\pi/2} f(x,y) \;dx\,dy[/tex]
 
  • #10
HallsofIvy
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You are surely not intended to use "Fresnel" integrals. That's the point of changing the order of integration. If you integrate with respect to a different variable first you will introduce another "x" into the integral and can substitute for the "x2".

The problem is
[tex]\displaystyle{\int_{z=0}^{1}\int_{y=0}^{1}\int_{x= 2y}^{2 } \frac{2\cos (x^2)}{\sqrt{z}}\;dx dy dz}[/tex]
Notice that I have added "z= ", "y= ", and "x= " to the limits of integration. I think that helps remember what you are doing.

I think we can ignore the "dz" and just swap x and y. y goes from 0 to 1 and, for each y, x goes from x= 2y to x= 2. I recommend you draw a picture of the region described by that: it is a triangle with vertices at (0,0), (2, 0), and (2, 1). Now, to reverse the order of integration, since "dx" will become the "outer integral" (still ignoring "dz") its limits must be constants. Look at your picture- what is the smallest value x takes on in that region? What is the largest value x takes on in that region? Those will be your limits of integration on the x integral. Now, for each x, what are the smallest and largest values y takes on? Imagine a vertical line crossing the region. What are the y values of the endpoints as functions of x? Those will be the limits of integration for the y- integral.

The first integral, with respect to y, is now very easy and, since the limits of integration now depend on x, will introduce an "x" into the integrand that
lets you substitute for x2 in the second integral.
 

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