Double Integral via Appropriate Change of Variables

[PhysicsWow]

Summary:: Calculate a double integral via appropriate change of variables in R^2

Suppose I have f(x,y)=sqrt(y^12 + 1). I need to integrate y from (x)^(1/11) to 1 and x from 0 to 1. The inner integral is in y and outer in x. How do I calculate integration(f(x,y)dxdy) ?

My Approach: I know that I have to change the variables to u-v plane from x,y, and I have seen the boundaries of the required region. But I can't express x and y as a function in (u,v)? Can someone give an insight into what should u and v appropriately be? Also, could you explain a bit why they should be assumed so?

 

[SammyS]

Summary:: Calculate a double integral via appropriate change of variables in R^2

Suppose I have f(x,y)=sqrt(y^12 + 1). I need to integrate y from (x)^(1/11) to 1 and x from 0 to 1. The inner integral is in y and outer in x. How do I calculate integration(f(x,y)dxdy) ?

My Approach: I know that I have to change the variables to u-v plane from x,y, and I have seen the boundaries of the required region. But I can't express x and y as a function in (u,v)? Can someone give an insight into what should u and v appropriately be? Also, could you explain a bit why they should be assumed so?

Hello, @PhysicsWow .


Is this an exercise in which you are required to change the variables to u-v plane from x,y ?

If not, it appears to me that the integration can be nicely accomplished simply by changing the order of integration.

 

[BvU]

Hello @PhysicsWow , !

Am I reading this right ? You want to calculate
$$\int_0^1 \int_{x^{1/11}}^1 \sqrt{ y^{12}+ 1\;} \, dy\,dx $$which is an integral of a function of y only

simply by changing the order of integration.

I don't see that. The bounds for y are a function of x. Am I missing something ?

 

[SammyS]

... simply by changing the order of integration.

I don't see that. The bounds for y are a function of x. Am I missing something ?

The bounds can easily be changed to have $x$ be a function of $y$ .