Finding the volume using double integrals

In summary, the integral given is set up correctly for the region enclosed by y = sqrt(x), x = 1, y = 2, and the intersection of those lines and the curve, which looks like a triangle. However, the second integral does not have a simple antiderivative and must be solved using numerical methods.
  • #1
farmd684
49
0

Homework Statement


R be the region enclosed by x=y^2,x=y=1 and y=2 hence evaluate
[tex]]

\iint e^{x/y^{2}} da
[/tex]

Homework Equations


The Attempt at a Solution



i have setup the integral as
[tex]
\int_{1}^{2} \int_{1}^{y^2} e^{x/y^{2}} dxdy
[/tex]
so far i got here
[tex]
\int_{1}^{2} ey^{2}dy - \int_{1}^{2} y^{2} e^{1/y^2} dy
[/tex]


I want to know that if i m right upto this point.
Thanks
 
Last edited:
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  • #2
anyone please :|
 
  • #3
Where did you get the function
[tex]
e^{x/y^2}?
[/tex]

I'm confused a little by your latex formatting - use { and not ( for enclosing the things in the super/subscript.

Are you sure that's the right function? If so, what you've shown in the post looks correct. Unfortunately, I don't think you're going to get a simple answer (just in terms of elementary functions) from the integral
[tex]
\int_1^2 y^2e^{1/y^2}\textrm{d}\,y
[/tex]
 
  • #4
farmd684 said:

Homework Statement


R be the region enclosed by x=y^2,x=y=1 and y=2 hence evaluate

It isn't clear what region you are trying to describe. What are the boundaries of your region?
 
  • #5
tmccullough said:
Where did you get the function
[tex]
e^{x/y^2}?
[/tex]

I'm confused a little by your latex formatting - use { and not ( for enclosing the things in the super/subscript.

Are you sure that's the right function? If so, what you've shown in the post looks correct. Unfortunately, I don't think you're going to get a simple answer (just in terms of elementary functions) from the integral
[tex]
\int_1^2 y^2e^{1/y^2}\textrm{d}\,y
[/tex]

The term [tex]e^{x/y^2}[/tex] was given in the question i was asked to integrate between the region limits.Then could you please tell me how to get the solution of [tex]
\int_1^2 y^2e^{1/y^2}\textrm{d}\,y
[/tex]
Thanks for the latex info i m still learning.

LCKurtz said:
It isn't clear what region you are trying to describe. What are the boundaries of your region?

The region is actually in the first quadrant [tex] y=\sqrt{x}[/tex] ,x=1, y=1 and y=2 and the intersection of those lines and curve region looks like a triangle.
 
  • #6
farmd684 said:
The region is actually in the first quadrant [tex] y=\sqrt{x}[/tex] ,x=1, y=1 and y=2 and the intersection of those lines and curve region looks like a triangle.

y = sqrt(x) is a piece of a parabola, x = 1 is a vertical straight line, y = 1 and y = 2 are two horizontal straight lines. You have given 4 "sides" and are talking about a "triangular" area?? Is y = 1 one of the sides?
 
  • #7
i have uploaded the picture of the region the shaded region i m talking about.

Code:
http://i30.tinypic.com/2920l6a.png
 
  • #8
Jakk01 said:
1 is the upper limit, not [tex]y^2[/tex]

i know and i guess my integration limits are right :smile:
 
  • #9
farmd684 said:
i have uploaded the picture of the region the shaded region i m talking about.

Code:
http://i30.tinypic.com/2920l6a.png

So y = 1 is not a boundary of the region and shouldn't be used in the description. You might have described it as the region above y = sqrt(x), below y = 2 and to the right of x = 1.
 
  • #10
Sorry OP, I misread your question, ignore what I said.
 
  • #11
LCKurtz said:
So y = 1 is not a boundary of the region and shouldn't be used in the description. You might have described it as the region above y = sqrt(x), below y = 2 and to the right of x = 1.

i m using y=1 as a lower limit of my y limit. Well i want to know if my double integral setup is correct or not and solution too please.
 
  • #12
farmd684 said:
i have setup the integral as
[tex]
\int_{1}^{2} \int_{1}^{y^2} e^{x/y^{2}} dxdy
[/tex]
so far i got here
[tex]
\int_{1}^{2} ey^{2}dy - \int_{1}^{2} y^{2} e^{1/y^2} dy
[/tex]

farmd684 said:
i m using y=1 as a lower limit of my y limit. Well i want to know if my double integral setup is correct or not and solution too please.

Now that I understand the region, yes you have it set up correctly. And as tmccullough has already noted, you aren't going to be able to find a simple antiderivative for that second integral.
 
  • #13
LCKurtz said:
Now that I understand the region, yes you have it set up correctly. And as tmccullough has already noted, you aren't going to be able to find a simple antiderivative for that second integral.

can you tell me how to solve it or refer me to any topics from which i can learn how to deal with these type of problems.

Thanks
 
  • #14
farmd684 said:
can you tell me how to solve it or refer me to any topics from which i can learn how to deal with these type of problems.

Thanks

Well, you apparently know how to set them up. When you wind up with a definite integral you can't do in a nice closed form, you must use numerical methods to get a decimal answer as accurate as you need. You could use Simpson's rule or if you have access to a high level mathematics program like Maple or Mathematica it will give you an immediate decimal answer. There are also resources on the internet such as Wolfram Alpha you can use.
 
  • #15
thank you :smile:
 

What is the concept of finding volume using double integrals?

The concept of finding volume using double integrals is based on the idea of dividing a three-dimensional solid into infinitesimally small pieces and summing up their volumes. It involves using double integrals in the Cartesian coordinate system to calculate the volume of a solid bounded by a surface in the xy-plane and a curve in the z-direction.

What are the steps involved in finding volume using double integrals?

The steps involved in finding volume using double integrals are:

  1. Define the region of integration in the xy-plane.
  2. Set up the double integral by determining the limits of integration and the integrand.
  3. Integrate the inner integral with respect to x.
  4. Integrate the outer integral with respect to y.
  5. Evaluate the limits and solve the double integral to find the volume.

What are the common applications of finding volume using double integrals?

Finding volume using double integrals has various applications in science and engineering. It is used to determine the volume of irregularly shaped objects such as planets, fluids, and biomolecules. It is also used in calculating the mass and density of an object, as well as in solving optimization problems in physics and economics.

What are the limitations of using double integrals to find volume?

One limitation of using double integrals to find volume is that it can only be applied to solids that can be described by a continuous function in the Cartesian coordinate system. It also becomes increasingly complex to solve when dealing with more complex shapes and regions of integration. Other methods, such as triple integrals, may be required for more complicated volumes.

How is finding volume using double integrals related to triple integrals?

Finding volume using double integrals is closely related to triple integrals as both involve calculating the volume of a three-dimensional solid. However, double integrals are used when the solid is bounded by a surface in the xy-plane and a curve in the z-direction, while triple integrals are used when the solid is bounded by three surfaces in the x, y, and z directions. Triple integrals are a more general form of double integrals and can be used in more complex situations.

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