Finding the volume using double integrals

  • Thread starter farmd684
  • Start date
  • #1
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:

Answers and Replies

  • #2
49
0
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
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770

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
49
0
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.

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
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
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
49
0
i have uploaded the picture of the region the shaded region i m talking about.

Code:
http://i30.tinypic.com/2920l6a.png
 
  • #8
49
0
1 is the upper limit, not [tex]y^2[/tex]

i know and i guess my integration limits are right :smile:
 
  • #9
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
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
9
0
Sorry OP, I misread your question, ignore what I said.
 
  • #11
49
0
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 wanna know if my double integral setup is correct or not and solution too please.
 
  • #12
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
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 m using y=1 as a lower limit of my y limit. Well i wanna 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
49
0
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
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
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
49
0
thank you :smile:
 

Related Threads on Finding the volume using double integrals

Replies
2
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
6K
Replies
1
Views
8K
Replies
3
Views
12K
Replies
2
Views
2K
Replies
3
Views
898
  • Last Post
Replies
2
Views
1K
Replies
5
Views
1K
Top