Double integrals - do areas cancel?

[dyn]

If i do a double integral of 1.dxdy to find an area of an odd function eg. y=x from +a to -a i get zero because the area below the x-axis cancels with the area above the x-axis.
If i do a double integral of a circle centred at the origin i get the area to be πr² ; so why doesn't the area below the x-axis cancel the area above the x-axis ?
Similarly with a volume integral of a sphere why doesn't any of the volume cancel another part of the volume ?
Thanks

 

[Office_Shredder]

I think if you integrate 1 over the pair of triangles between the x-axis and the line y=x you get the area of the two triangles. This is not the same as integrating the function y=x between -a and a.

 

[dyn]

If i integrate in 1-D the function y=x between x=a and x= -a i get zero, which is the same answer if i evaluate the area as a double integral

 

[Delta2]

If i integrate in 1-D the function y=x between x=a and x= -a i get zero, which is the same answer if i evaluate the area as a double integral

Could you show us your calculation of the double integral? According to me (and also to post #2) you don't get zero, you get the total area of those two triangles. You just can't integrate a strictly positive function (the constant function with value 1) and get zero as result...

 

[Stephen Tashi]

If i integrate in 1-D the function y=x between x=a and x= -a i get zero, which is the same answer if i evaluate the area as a double integral

This is a question of vocabulary.

The standard interpretation for finding the integral of a function of one variable "between" two numbers is to treat the first numer mentioned as the lower limit of integration and the second number as the upper limit of integration. We can regard this convention as saying we treat all lengths on the x-axis to be non-negative.

When a text poses a problem like "Find the area of $f(x,y) =...$ over the square with opposite corners $(-1,-1), (1,1)$", it is usually understood that the student is to treat all parts of the area of the square as positive, even though there can exists applications where, for example, the area of the sub-square with corners $(0,-1), (1,0)$ might be interpreted as representing some physical quantity that is negative.

Saying we evaluate a function "over an area" as a double integral suggests that we want to use the usual textbook interpretation of that task. Saying we "find the area" of some shape also suggests that we want to set up our double integrals so all parts of the region produce non-negative answers for their area.

For example, the integrand of a double integral may have a Jacobian determinant as a factor. This factor may be a function that can take on negative values. When a problem asks for finding the integral of a function "over an area", the convention is to put absolute values signs around the Jacobian determinant.

The usual interpretation of "finding the area" of a region in the plane is a special case of finding the area of a function "over an area". We use the constant function $f(x,y)=1$.

(The use of lowercase "i" for the personal pronoun "I" makes typography like "a i" confusing. I assume we aren't talking about imaginary numbers. )

 

[Mark44]

This is a question of vocabulary.

I agree. There's a difference between "find the integral of f(x) between x = -a and x = b" versus "find the area of the triangular regions bounded by the graph of y = x and the lines x = -a and x = a"

If i integrate in 1-D the function y=x between x=a and x= -a i get zero, which is the same answer if i evaluate the area as a double integral

Forget double integration for the time being, as I believe your confusion is between a) the integral of a function, and b) the area of a particular region or regions.

If we simply integrate, we get $\int_{-a}^a x dx = \frac 1 2 x^2|_{-a}^a = \frac 1 2 (a^2 - (-a)^2) = 0$
On the other hand, if the problem asks for the areas of the two triangular regions, the integrands will be different on the two halves of the interval.
$\int_{-a}^0 (0 - x)dx + \int_0^a x dx = \frac 1 2 \left(-x^2|_{-a}^0 + x^2|_0^a\right) = \frac 1 2 \left(0 - (-a^2) + a^2 - 0\right) = \frac 1 2 \left( a^2 + a^2\right) = a^2$
In the first integral just above, we have to take the height of the typical area element as positive, so each height will be (0 - y), or (0 - x), since y = x.

 

[dyn]

If i have a graph of y= x and integrate in 1-D from x= -a to x=a then i get zero because " the negative area cancels the positive area"
If i do a double integral of a circle from from x = -a to x= +a i get πa² and if i do a volume integral of a sphere from x = -a to x= +a i get (4/3)πa³
Why do areas cancel in the 1st case but the area in the 2nd case and volume in the 3rd case not cancel ?

 

[fresh_42]

Why do areas cancel in the 1st case but the area in the 2nd case and volume in the 3rd case not cancel ?

Areas and volumes are orientated quantities: surrounding them clockwise results in the opposite sign as counter clockwise. If you keep the orientation, i.e. the sign, then volumes can cancel each other. This happens to $\int_{-a}^a x\,dx$ since you walk around the two areas in different orientations (along the positive x-axis). If you set up the double integral correctly (suited parametrization), this shouldn't happen in that case. However, you denied to show us your computation although being asked for.

If you are interested in the physical area or volume, you have to add the absolute values of all subareas or subvoluminae, or carefully parameterize the way around the area. Roughly speaking it is $dx\wedge dy$ versus $dx(t) dy(t)$.

 

[dyn]

Is it because area integrals such as ∫∫ 1.dxdy and volume integrals such as ∫∫∫ 1. dxdydz have to be positive by definition ?
Is an area element dxdy or volume element dxdydz positive everywhere ? What about in the 2nd quadrant for example where x is negative and y is positive. Is dxdy still positive ?

 

[fresh_42]

Is it because area integrals such as ∫∫ 1.dxdy and volume integrals such as ∫∫∫ 1. dxdydz have to be positive by definition ?

Depends on how you use them. They are positive in common language, signed in mathematics. You could as well explain it with Stokes' theorem: 'The integral (...) over a closed path (...) is zero.' This means: you need to get a different sign if you only walk half the way, because then forward and backwards makes a difference.

Is an area element dxdy or volume element dxdydz positive everywhere ? What about in the 2nd quadrant for example where x is negative and y is positive. Is dxdy still positive ?

This is too general to answer. $dxdy=dydx$ in the cases we discussed here. But $dx\wedge dy= -dy \wedge dx$. If you want to understand the sign, you need either accept orientation or learn about differential forms.

 

[dyn]

Areas and volumes are orientated quantities: surrounding them clockwise results in the opposite sign as counter clockwise. If you keep the orientation, i.e. the sign, then volumes can cancel each other. This happens to $\int_{-a}^a x\,dx$ since you walk around the two areas in different orientations (along the positive x-axis). If you set up the double integral correctly (suited parametrization), this shouldn't happen in that case. However, you denied to show us your computation although being asked for.

If you are interested in the physical area or volume, you have to add the absolute values of all subareas or subvoluminae, or carefully parameterize the way around the area. Roughly speaking it is $dx\wedge dy$ versus $dx(t) dy(t)$.

If i do the double integral of y=x from -a to +a then i have to split the integral into 2 halfs and i don't get zero , i get a²

The RHS is ∫∫ dxdy with y limits from 0 to a and x limits from 0 to y giving a²/2

The LHS is ∫∫ dxdy with y limits from -a to 0 and x limits from 0 to y giving a²/2

Adding these together gives a²

 

[Mark44]

If i have a graph of y= x and integrate in 1-D from x= -a to x=a then i get zero because " the negative area cancels the positive area"

I already discussed this in post #6. Did you read my post? There can be a difference between the integral of a function over some interval and the notion of the area between the graph of the function and the horizontal axis. The difference is present when the graph lies on different sides of the horizontal axis. You are treating the integral and the area as being synonymous -- they aren't, as @fresh_42 is also pointing out.

 

[fresh_42]

I already discussed this in post #6. Did you read my post?

Yes, I did. I only have the impression that we do not get through to @dyn, so I gave it a shot. My main goal was to speak about the concept of orientation, since this is a property which runs through entire physics.

 

[Mark44]

Yes, I did.

My question was to @dyn.

 

[dyn]

My question was to @dyn.

I have read all the posts but in my experience so far orientation does not come into double or triple integrals unless they are surface integrals

 

[Mark44]

I have read all the posts but in my experience so far orientation does not come into double or triple integrals unless they are surface integrals

The point you seem to be missing is the difference between 1) the integral of a function over an interval (single integral) or over a region in the plane (double integral), and 2) calculating an area or volume using integration.

From post #1:

If i do a double integral of 1.dxdy to find an area of an odd function eg. y=x from +a to -a i get zero because the area below the x-axis cancels with the area above the x-axis.

No. The area below the line is positive.
Again, it doesn't matter whether you're doing a single integral or a double integral. You are confusing the area between the graph of a function and the horizontal axis with the integral of the same function over the same interval.
Here are two problems that show the difference.
1) Calculate $\int_0^{2\pi} \sin(x) dx$
Answer: $\int_0^{2\pi} \sin(x) dx = 0$
2) Find the area between the graph of $y = \sin(x)$ and the x-axis, on the interval $[0, 2\pi]$
Answer: $\int_0^{\pi} \sin(x) dx + \int_\pi^{2\pi} (0 - \sin(x))dx = 4$
I leave the grunt work of calculating the integrals to you.

If i do a double integral of a circle centred at the origin i get the area to be πr2 ; so why doesn't the area below the x-axis cancel the area above the x-axis ?

An integral can produce a negative value, but areas aren't negative. If your integration is done on x first and then y, you are calculating horizontal slices going from left to right, and then summing them going from bottom to top. Doing so will give you a positive number.
If your integration is done on y first and then x, you are calculating vertical slices going from bottom to top, and then summing them going from left to right. Doing things this way will also give you a positive number, the same as if you had integrated in the other order.

 

[jtbell]

The RHS is ∫∫ dxdy with y limits from 0 to a and x limits from 0 to y giving a²/2

OK.

The LHS is ∫∫ dxdy with y limits from -a to 0 and x limits from 0 to y giving a²/2

I would write the x limits as going from y to 0: from lower x to higher x. y is negative for this part of the area.
$$\int_{-a}^0 \int_y^0 dx\,dy = -\int_{-a}^0 y\,dy = \frac {a^2} 2$$

 

[Mark44]

I would write the x limits as going from y to 0: from lower x to higher x. y is negative for this part of the area. $$\int_{-a}^0 \int_y^0 dx\,dy = -\int_{-a}^0 y\,dy = \frac {a^2} 2$$

This ties into what I said in the next-to-last paragraph of post #16.

 

[dyn]

OK.

I would write the x limits as going from y to 0: from lower x to higher x. y is negative for this part of the area.
$$\int_{-a}^0 \int_y^0 dx\,dy = -\int_{-a}^0 y\,dy = \frac {a^2} 2$$

So is the order of the limits in a double or even triple integral always in the order of increasing x and increasing y (and increasing z in a triple integral ) ?

 

[Stephen Tashi]

Is it because area integrals such as ∫∫ 1.dxdy and volume integrals such as ∫∫∫ 1. dxdydz have to be positive by definition ?

By convention, a problem that asks you to find the area or volume of a shape is asking you to treat all parts of that shape as non-negative quantities.

As to whether "$\int \int \int dxdydz$" is non-negative, that notation doesn't indicate a specific definite integral until the limits of integration are given. If you are using the notation "$\int \int \int dxdydz$" to abbreviate the words "The volume of some shape" then, yes, all parts of that volume are defined to have non-negative volume and you are expected to arrange the calculation so things work out that way.

Is an area element dxdy or volume element dxdydz positive everywhere ?

I'd say yes - at least it's non-negative. How people think about symbolism such as "dxdydz" varies. If you think of a finite Riemann sum approximating an integral and of "dxdydz" denote the volume of the rectangular solids involved in the approximation then dxdydz represents a non-negative number. This due to the adjective "volume". It is possible to have integrals that include the symbolism "dxdydz" whose physical interpretation is not an integration over a volume.

What about in the 2nd quadrant for example where x is negative and y is positive. Is dxdy still positive ?

If you are thinking of $dx$ as a finite quantity, the fact that $x < 0$ does not prevent $dx$ from being positive. For example $dx = (x + dx) - x = (-2) - (-3) = 1$ when $x = -3, dx = 1$.

 

[Stephen Tashi]

So is the order of the limits in a double or even triple integral always in the order of increasing x and increasing y (and increasing z in a triple integral ) ?

You aren't paying attention to the fact that terminology like "double integral" does not define a specific mathematical problem. The topic of the current thread is focused on working problems that involve areas and volumes. There can be problems that do not involve areas and volumes and multiple integrals may be used to solve such problems. So: No, there is no general rule that says the limits used in multiple integrals must set the upper limit to be greater than the lower limit - but, Yes, in problems involving areas and volumes, the general idea is that you arrange things so any factor in the integrand representing an area or volume is non-negative.

To say we keep factors representing areas and volume non-negative by making the order of the limits increasing in x and y is a little too simplistic. Some limits in double integrals may be functions of x and y. The limits are not always constants.

 

[dyn]

I was under the impression that volumes can be calculated in either one of 2 ways ; ∫∫ z(x,y) dxdy or ∫∫∫ dxdydz

I have tried both ways for the example of a cube of sides 2a lying below the xy plane with one face lying in the xy plane. The double integral gives an answer of -8a³ and the triple integral gives an answer of 8a³. So are the 2 integrals always equal in magnitude but can differ in sign ? If the double integral can give negative answers then for a volume lying above and below the xy plane we could get back to zero volumes ; so is the triple integral a better way of obtaining a positive volume

 

[fresh_42]

If the double integral can give negative answers then for a volume lying above and below the xy plane we could get back to zero volumes ; so is the triple integral a better way of obtaining a positive volume

These are two different things, i.e. it won't save your day. Volumes are oriented as well, and so are integrals. If you want to calculate a physical volume, them you have to make sure that the orientation doesn't change on the way, the same as with areas and double integrals. Integrals are certain operators. They can only be used to calculate areas and volumes as you can use a hammer to hang a picture. It is a tool, but as hammers weren't invented for pictures, so weren't integrals set up to calculate areas and volumes.

 

[dyn]

Thank you everyone. It seems as though everything i have learned is wrong

 

[sysprog]

Thank you everyone. It seems as though everything i have learned is wrong

Well, not entirely, methinks:

This is a question of vocabulary.

$-$ e.g. Jello^® rings present "volume" and Jello^® ring mold interiors present "capacity".

The two quantities could be written (in some insular argot) as $n\mathtt{cm}^3$ and $-n\mathtt{cm}^3$, respectively.