Proof of The volume under surfaces formula

  • I
  • Thread starter TheoEndre
  • Start date
  • #1
42
3
Hello everyone,
Is there a proof that takes us from the sum idea of the volume:
$$\sum_{i=1}^m \sum_{j=1}^n f(x_i,y_j) \Delta x \Delta y$$
To the integral idea:
$$\iint_R f(x,y) dxdy$$
Or something that relates the volume to the integral just like The Fundamental Theorem of Calculus?



 

Answers and Replies

  • #2
verty
Homework Helper
2,182
198
I recommend thinking of each integral or each sum as being entirely separate. The outer sum in the first case, and the outer integral in the second case, sees just an expression, the result of the inner sum or integral. Each behaves entirely normally so there is nothing to prove (surely there is nothing to prove).
 
  • #3
42
3
I recommend thinking of each integral or each sum as being entirely separate. The outer sum in the first case, and the outer integral in the second case, sees just an expression, the result of the inner sum or integral. Each behaves entirely normally so there is nothing to prove (surely there is nothing to prove).
This is weird. So there is no need for proof for this thing... I thought they'll use "Generalized fundamental theorem of calculus" or something like that.
But It seems like this is not the case. By the way, is there something like "Generalized fundamental theorem of calculus"?
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
This is weird. So there is no need for proof for this thing... I thought they'll use "Generalized fundamental theorem of calculus" or something like that.
But It seems like this is not the case. By the way, is there something like "Generalized fundamental theorem of calculus"?

The definition of the integral is the limit of the appropriate sum; in this case, as ##\Delta x, \Delta y \rightarrow 0##.

That said, there's every likelihood that someone will come along on this thread with a proof!
 
  • #5
42
3
The definition of the integral is the limit of the appropriate sum; in this case, as ##\Delta x, \Delta y \rightarrow 0##.

That said, there's every likelihood that someone will come along on this thread with a proof!
I know that, but the relation to antiderivative is not clear from the limit of the sum (I think so).
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
I know that, but the relation to antiderivative is not clear from the limit of the sum (I think so).

How does that relate to your original question?
 
  • #7
42
3
How does that relate to your original question?
Sorry about that, I always forget to mention antiderivative when taking about integrals. But, this is what I want to know the answer to: How does the volume relate to the antiderivative of the function ##f(x,y)##?
Really sorry about that.
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
Sorry about that, I always forget to mention antiderivative when taking about integrals. But, this is what I want to know the answer to: How does the volume relate to the antiderivative of the function ##f(x,y)##?
Really sorry about that.

Volume implies a definite integral. And a double integral is just one integral after another. To do an example:

##\int_0^1 \int_0^2 2xy^2 dx dy = \int_0^1 (\int_0^2 2xy^2 dx) dy = \int_0^1 (x^2y^2|_0^2) dy = \int_0^1 4y^2 dy \dots##

So, you need to break the antidifferentiation into two steps as above.
 
  • #9
42
3
Volume implies a definite integral. And a double integral is just one integral after another. To do an example:

##\int_0^1 \int_0^2 2xy^2 dx dy = \int_0^1 (\int_0^2 2xy^2 dx) dy = \int_0^1 (x^2y^2|_0^2) dy = \int_0^1 4y^2 dy \dots##

So, you need to break the antidifferentiation into two steps as above.
My question is to know why taking two antiderivatives relates to volume? The fundamental theorem of calculus (The first and second) and their proofs made the relation between antiderivative and area under a curve makes a lot of sense, I wondered if there was something like that relates double antiderivatives to the volume under a surface
 
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
My question is to know why taking two antiderivatives relates to volume? The fundamental theorem of calculus (The first and second) and their proofs made the relation between antiderivative and area under a curve makes a lot of sense, I wondered if there was something like that relates double antiderivatives to the volume under a surface

It's a fairly logical extension of the single-variable case. See, for example:

http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx
 
  • #12
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
I see where this is going, but I'm a bit disappointed because I really like rigorous proofs of things in math.

Can you state the theorem that you would like to find a proof of?
 
  • #13
42
3
Can you state the theorem that you would like to find a proof of?
Let: ##f(x,y)=z## be a continuous surface on the rectangular region ##R##. If The volume under the surface above that region is given by:
$$V= \iint_R f(x,y) dxdy$$
Prove that:
  1. $$\frac {\partial^2 V} {\partial y \partial x} =f(x,y)$$ $$And$$ $$\frac {\partial^2 V} {\partial x \partial y}=f(x,y)$$
  2. Prove the evaluation ( just like ##\int_a^b f(x) dx = F(b)-F(a)##, I don't know how to state it here).
I hope this is good as a statement
 
  • #14
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
Let: ##f(x,y)=z## be a continuous surface on the rectangular region ##R##. If The volume under the surface above that region is given by:
$$V= \iint_R f(x,y) dxdy$$
Prove that:
  1. $$\frac {\partial^2 V} {\partial y \partial x} =f(x,y)$$ $$And$$ $$\frac {\partial^2 V} {\partial x \partial y}=f(x,y)$$
  2. Prove the evaluation ( just like ##\int_a^b f(x) dx = F(b)-F(a)##, I don't know how to state it here).
I hope this is good as a statement

Here ##V## is a number (independent of ##x, y##), so cannot be differentiated. ##V## depends on ##R##.
 
  • #15
42
3
Here ##V## is a number (independent of ##x, y##), so cannot be differentiated. ##V## depends on ##R##.
How about:
$$V= \int_{t_2=c_2}^{t_2=y} \int_{t_1=c_1}^{t_1=x} f(t_1,t_2) dt_1 dt_2$$
Where ##t_1 , t_2## are dummy variables and ##c_1 , c_2## are constants
 
  • #16
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
How about:
$$V= \int_{t_2=c_2}^{t_2=y} \int_{t_1=c_1}^{t_1=x} f(t_1,t_2) dt_1 dt_2$$
Where ##t_1 , t_2## are dummy variables and ##c_1 , c_2## are constants

What do you think? Can you extend what you know from the FTC?
 
  • #17
42
3
What do you think? Can you extend what you know from the FTC?
I tried my best. Tried to construct an average area ##A_{avg}## where: ##A_{avg_x} . \Delta y=V## and ##A_{avg_y} . \Delta x=V## but nothing.
Tried to construct an average hight ##f_{avg}## where: ##f_{avg} . \Delta x \Delta y=V##, but nothing.
Tried to use the definition of the partial derivatives on ##V## And used what I constructed ##(A_{avg_x} . \Delta y=V) , (A_{avg_y} . \Delta x=V) , (f_{avg} . \Delta x \Delta y=V)## but nothing. Maybe because of wrong derivations.
 
  • #18
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
Well, you know that:

##\int_{c_1}^{x} f(t_1, t_2) dt_1 = F(x, t_2) - F(c_1, t_2)##

Where ##F## is the antiderivative of ##f## with respect to ##x##.
 
  • #19
42
3
Well, you know that:

##\int_{c_1}^{x} f(t_1, t_2) dt_1 = F(x, t_2) - F(c_1, t_2)##

Where ##F## is the antiderivative of ##f## with respect to ##x##.
Fair enough. So by taking the second integral, we get ##F(x,y)-F(c_1,c_2)##. I see how this is done, maybe I was overthinking it. But still, I feel like it could be proven using the definition of the partial derivatives and something else along with it. Just like when we used the definition of derivatives on ##F(x)## along with the Mean Value Theorem for definite integrals to prove the connection.
 
  • #20
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
Fair enough. So by taking the second integral, we get ##F(x,y)-F(c_1,c_2)##. I see how this is done, maybe I was overthinking it. But still, I feel like it could be proven using the definition of the partial derivatives and something else along with it. Just like when we used the definition of derivatives on ##F(x)## along with the Mean Value Theorem for definite integrals to prove the connection.

You're just going to get a bit of a mess with all the various antiderivatives. There's nothing really new theoretically with the double integral - it's just a bit more complicated.
 
  • #21
42
3
You're just going to get a bit of a mess with all the various antiderivatives. There's nothing really new theoretically with the double integral - it's just a bit more complicated.
I see... Thanks for clearing thinks to me.
 
  • #22
mathwonk
Science Advisor
Homework Helper
11,307
1,522
the derivative of a double integral, i.e. of a volume, wrt one of the variables, is like asking for the rate of change of the volume of a loaf of bread, as we move the slicer along the loaf. the answer is it equals the area of a slice of the bread. i.e. the derivative of a moving volume function, as we move the plane that is slicing the solid involved, is just the area of a slice. thus the derivative of a double integral wrt say x, at a given point, is the single integral with respect to y, at that fixed value of x. then the integral of that area function wrt y, is of course the value of the height function at (x and) y. i.e. the derivative wrt the first variable is an area, and then the derivative of that area wrt the second variable is the height, as usual in one variable calculus.
 
  • Like
Likes member 587159
  • #23
42
3
the derivative of a double integral, i.e. of a volume, wrt one of the variables, is like asking for the rate of change of the volume of a loaf of bread, as we move the slicer along the loaf. the answer is it equals the area of a slice of the bread. i.e. the derivative of a moving volume function, as we move the plane that is slicing the solid involved, is just the area of a slice. thus the derivative of a double integral wrt say x, at a given point, is the single integral with respect to y, at that fixed value of x. then the integral of that area function wrt y, is of course the value of the height function at (x and) y. i.e. the derivative wrt the first variable is an area, and then the derivative of that area wrt the second variable is the height, as usual in one variable calculus.
Awesome picture for what is happening when taking partial derivatives... thanks for that.
 
  • #24
mathwonk
Science Advisor
Homework Helper
11,307
1,522
you are welcome. after seeing this i finally understood formulas like that for the integral of a solid of revolution. e.g. if revolving y = f(x) about the x axis, the slices perpendicular to the x axis are circles with radius equal to f(x), so the volume formula is the integral of π (f(x))^2 dx. In general the area of a slice perpendicular to the x axis, is the integral of f(x,y) wrt y with x held constant, so the volume is the integral of "area of the slice at x" dx, or the integral of "the integral of f(x,y)dy" dx.
 
  • #25
42
3
you are welcome. after seeing this i finally understood formulas like that for the integral of a solid of revolution. e.g. if revolving y = f(x) about the x axis, the slices perpendicular to the x axis are circles with radius equal to f(x), so the volume formula is the integral of π (f(x))^2 dx. In general the area of a slice perpendicular to the x axis, is the integral of f(x,y) wrt y with x held constant, so the volume is the integral of "area of the slice at x" dx, or the integral of "the integral of f(x,y)dy" dx.
Wow... Indeed, it makes more sense if you think of it that way. man... I really wish I can visualize things like this... It seems I have a lot of things to experience in order to reach this level. By the way, thanks for the insight, really helped.
 
  • #26
mathwonk
Science Advisor
Homework Helper
11,307
1,522
thank you again. this came to me slowly over many years, as I taught calculus from 1970 to 2010, and asked myself what was going on every year anew. just keep trying, and accept that it is a long process. With every iteration it gets clearer, but nothing becomes obvious after one take., so i advise not to expect that one pass can offer good understanding. i am 75 years old and still reading basic books on my own specialty, hoping to gain more insight before i lose the ability to read and understand. To be clear, after 40 years of teaching and research, I am still trying to add to my understanding of basic facts about my own research specialty. Why not? It's really fun!
 
  • #27
42
3
thank you again. this came to me slowly over many years, as I taught calculus from 1970 to 2010, and asked myself what was going on every year anew. just keep trying, and accept that it is a long process. With every iteration it gets clearer, but nothing becomes obvious after one take., so i advise not to expect that one pass can offer good understanding. i am 75 years old and still reading basic books on my own specialty, hoping to gain more insight before i lose the ability to read and understand. To be clear, after 40 years of teaching and research, I am still trying to add to my understanding of basic facts about my own research specialty. Why not? It's really fun!
Thank you very much for this great advice, I always get depressed if I don't understand something quickly but your advice made me a bit relieved :biggrin:
 

Related Threads on Proof of The volume under surfaces formula

  • Last Post
Replies
1
Views
6K
Replies
9
Views
20K
Replies
4
Views
4K
Replies
8
Views
917
Replies
7
Views
3K
  • Last Post
4
Replies
109
Views
117K
Replies
3
Views
25K
  • Last Post
Replies
24
Views
4K
  • Last Post
Replies
2
Views
8K
Replies
1
Views
3K
Top