Register to reply

Vector Calculus: Area and Mass of a Region

by Kushwoho44
Tags: calculus, mass, region, vector
Share this thread:
Dec1-12, 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 xy-plane 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 xy-plane 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 xy-plane 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.

Phys.Org News Partner Science news on
Apple to unveil 'iWatch' on September 9
NASA deep-space rocket, SLS, to launch in 2018
Study examines 13,000-year-old nanodiamonds from multiple locations across three continents
Dec1-12, 08:44 AM
Sci Advisor
PF Gold
P: 39,542
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.
Dec3-12, 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.

Register to reply

Related Discussions
Surface area of a hemisphere w/ vector calculus Calculus & Beyond Homework 1
Surface area of a hemisphere w/ vector calculus Calculus & Beyond Homework 2
Pre-Calculus - Regarding finding the area of a certain region Introductory Physics Homework 4
HELP:Calculus (area region) Introductory Physics Homework 5