# Finding the volume using double integrals

## Homework Statement

R be the region enclosed by x=y^2,x=y=1 and y=2 hence evaluate
$$] \iint e^{x/y^{2}} da$$

## The Attempt at a Solution

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

I want to know that if i m right upto this point.
Thanks

Last edited:

Where did you get the function
$$e^{x/y^2}?$$

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
$$\int_1^2 y^2e^{1/y^2}\textrm{d}\,y$$

LCKurtz
Homework Helper
Gold Member

## 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?

Where did you get the function
$$e^{x/y^2}?$$

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
$$\int_1^2 y^2e^{1/y^2}\textrm{d}\,y$$

The term $$e^{x/y^2}$$ 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 $$\int_1^2 y^2e^{1/y^2}\textrm{d}\,y$$
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 $$y=\sqrt{x}$$ ,x=1, y=1 and y=2 and the intersection of those lines and curve region looks like a triangle.

LCKurtz
Homework Helper
Gold Member
The region is actually in the first quadrant $$y=\sqrt{x}$$ ,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?

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

1 is the upper limit, not $$y^2$$

i know and i guess my integration limits are right LCKurtz
Homework Helper
Gold Member

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.

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.

LCKurtz
Homework Helper
Gold Member
i have setup the integral as
$$\int_{1}^{2} \int_{1}^{y^2} e^{x/y^{2}} dxdy$$
so far i got here
$$\int_{1}^{2} ey^{2}dy - \int_{1}^{2} y^{2} e^{1/y^2} dy$$

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.

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

LCKurtz
thank you 