
#1
Dec112, 02:40 AM

P: 10

Hi y'all.
Here is exactly what is stated on the theory page of my book: Example: Area of a Region The area of a region R in the xyplane corresponds to the case where f(x,y)=1. Area of R= ∫∫dR Example: Mass of a Region The mass of a region R in the xyplane with mass density per unit area ρ(x,y) is given by: Mass of R= ∫∫ρdR I'm not at all understanding this first part of the theory, why is it that the area of the region R in the xyplane is the case f(x,y)=1 and how did they obtain that express for the area R? All help is immensely appreciated as I'm tearing my hair out over this. Thanks! 



#2
Dec112, 08:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,885

What do you understand about integrals? Most people are first introduced to integrals in the for [itex]\int_a^b f(x)dx[/itex] where it is defined as "the area of the region bounded by the graphs of y= f(x), y= 0, x= a, and x= b".
Once you start dealing with double integrals, since [itex]f(x)= \int_0^{f(x)} dy[/itex], it is easy to see that we can write that integral, and so that area, as [itex]\int_a^b\int_0^{f(x)} dy dx[/itex]. From that we can see that "dxdy" acts as a "differential of area". That is, we find the area of any region by integrating [itex]\int\int dx dy[/itex] over that region. Similarly, in three dimensions, we can find the volume of a region by integrating [itex]\int\int\int dxdydz[/itex] over that region. Another way of reaching the same idea is to divide the region into small rectangles with sides parallel to the x and y axes and identifying the lengths of the sides as "[itex]\Delta x[/itex]" and "[itex]\Delta y[/itex]". Of course, the area of each such rectangle is [itex]\Delta x\Delta y[/itex] and the area of the whole region can be approximated by [itex]\sum \Delta x\Delta y[/itex]. "Approximate" because some of the region, near the bounds, will not fit neatly into those rectangles. But we can make it exact by taking the limit as the size of [itex]\Delta x[/itex] and [itex]\Delta y[/itex] go to 0. Of course, I have no idea what method your texts or courses used to introduce the double integral so I cannot be more precise. 



#3
Dec312, 09:01 PM

P: 10

Thanks for that, the stupid part I wasn't understanding was that f(x,y) maps onto z.
Makes perfect sense now. 


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