How do I find the definite integral in a changed coordinate system?

[RadiationX]

in all seriousness because of the hurricanes in florida i did not get to learn about finding integrals in the Y perspective. given that this is my question.

let R be the region in the first quadrant bounded by x=0,y=1,y=2, and
y= 4e^(-x^2). to my knowledge the integral needs to be set up so that we change perspectives make the y-axis the x-axis and the x-axis the y axis. so we cut the graph with an arbitray "horizantol" line. so this gives us the integral from x=1 to x=2 (former y values). here is my problem the first horizontal line that Y touches is Y=0 (former x value) and y = 4e^(-x^2). now since we changed our coordinate system don't we need to change the exponeltial function thus x=4e^(-y^2) and solve for Y?

 

[dextercioby]

Not really,u need to find
$$y^{-1}(x)$$

Daniel.

If you don't like it,then how about
$$y(x_{1})=1$$

$$y(x_{2})=2$$

The solutions will be your new integration limits...

 

[RadiationX]

isn't $$y^{-1}(x)$$ the same thing as x=4e^(-y^2) and solve for Y? is this not the inverse of Y also

 

[dextercioby]

Find x=x(y) and integrate between y_{1}=1 & y_{2}=2...

Daniel.

 

[RadiationX]

so is y=sqrt(ln4-lnx) my function to integrate?

 

[dextercioby]

Yes,it looks ugly,since it involves the function "erf"...

Daniel.

P.S.Really ugly indeed...

 

[RadiationX]

man that looks so ugly and WRONG i hope this is correct. so what i have now is this
integral from 1 to 2 of sqrt(ln4-lnx). this is the area you say

 

[dextercioby]

Yes,try to integrate:
$$\int_{1}^{2} \sqrt{\ln 4 -\ln x} \ dx$$

Daniel.

 

[RadiationX]

thank you. i'll try that

 

[RadiationX]

i used my calculator and got an are of approx 0.995121: this seems very small

 

[dextercioby]

I would't know what the final outcome would be,simply because it's pretty difficult to get...
Try to do it by hand,though.

Daniel.

 

[RadiationX]

i got it, the above is the correct solution. thank you