Finding the area of a double integral using dxdy instead of dydx

In summary, the conversation discusses the use of dydx as the area in solving a problem and the frustration of not being able to find a different solution. The conversation also explores different methods of solving the problem, including using a double integral and integrating with respect to x or y first. The correct solution is shown to be 1, and the reason behind the incorrect result of 2 is explained. It is concluded that extra effort is needed when using quantities like ## \int _a^b \ln x\mathrm{d}x##.
  • #1
ago01
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Homework Statement
Given The curve $$y = e^x$$ and the lines $$y = 0$$, $$x = 0$$, and $$x = ln 2$$ express the area as an iterated integral and solve.
Relevant Equations
N/A
I have the solution for this problem using dydx as the area. Worse yet, I cannot find another solution for it. Everyone seems to just magically pick dydx without thinking and naturally this is frustrating as learning the correct choice is 99.9% of the battle...

So, I was curious how one might go about solving it the other way. Here's a representation of the area:

1648327404567.png


So if we imagine going from left to right (for dx) then our x limits seem to go from $$y=e^x$$ to $$ln2$$ which implies $$x = lny$$ and $$x = ln2$$ for our lower and upper limits. Then it follows that y goes from $$0$$ to $$2$$. So:

$$\int^{2}_{0}\int^{ln2}_{lny}dxdy$$

But after solving this integral I get an area of 2, rather than the desired area of 1. I cannot figure out how my setup is wrong here. What am I missing?
 
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  • #2
The graph of ##y = e^x## does not look like your drawing of it.
 
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  • #3
I see. Yeah that would do it. e^x has y = 1 at x = 0. My graph is shoddy.

It's obvious now that if you go the x-direction you have two different functions. One being x= 0, and then it switches to y = e^x. In this case, you would need extra integration to cover that function switch, so dydx is really the only way to do it.
 
  • #4
You, of course, should know how to do it either way. So I wouldn't say using ##dy\,dx## is the only way to do it, but perhaps the preferable way to do it.

If you integrate with respect to ##x## first, you still really only need one integral since the other chunk of the area is a rectangle. You don't need to integrate to find its area.
 
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  • #5
The desired area is
[tex]
\int _0^{\ln 2} e^x \mathrm{d}x = e^{\ln 2} - e^0 = 1.
[/tex]
You can regard this area as a double integral as follows.
[tex]
\int\int _A \mathrm{d}y\mathrm{d}x = \int _0 ^{\ln 2} \int _0^1 \mathrm{d}y\mathrm{d}x + \int _0^{\ln 2} \int _1^{e^x}\mathrm{d}y\mathrm{d}x = \ln 2 + \int _{0}^{\ln 2}(e^x-1)\mathrm{d}x = 1.
[/tex]
Equivalently,
[tex]
\int\int _A \mathrm{d}x\mathrm{d}y = \int _0^1 \int_0^{\ln 2} \mathrm{d}x\mathrm{d}y + \int _1^2 \int _{\ln y}^{\ln 2} \mathrm{d}x\mathrm{d}y = \ln 2 + \int _1^2 (\ln 2 - \ln y)\mathrm{d}y = 1.
[/tex]
You need more effort to use quantities like ## \int _a^b \ln x\mathrm{d}x##.
That's why the invariance is often extremely useful, because in a certain order, computation could be significantly easier. Due to said invariance, one may conclude that
[tex]
\int _1^2 \ln y\mathrm{d}y = \ln 4 -1
[/tex]
without explicit computation. Sometimes one order might include an integral for which an antiderivative in terms of elementary functions is not known, but another order is trivial.

You might have thought that
[tex]
\int _0^2 \int _{\ln y} ^{\ln 2} \mathrm{d}x\mathrm{d}y = \int _0^1 \int _0^{\ln 2}\mathrm{d}x\mathrm{d}y + \int _1^2 \int _{\ln y}^{\ln 2} \mathrm{d}x\mathrm{d}y
[/tex]
but as we saw, this is not true. Why? Because on the left hand side, ##y## is allowed to range from ##0## and ##\ln y## is unbounded as ##y\to 0+## so the left hand side does not correspond to the area of interest.

The LHS allows you to compute the area under ##e^x## where ##-\infty < x < \ln 2##, which explains your result of ##2##.
 
Last edited:

What is a double integral?

A double integral is a type of integral in mathematics that involves integrating a function of two variables over a two-dimensional region. It is often used to find the volume or area under a surface.

What is the difference between dxdy and dydx in a double integral?

In a double integral, dxdy represents integrating with respect to the x variable first and then the y variable, while dydx represents integrating with respect to the y variable first and then the x variable. The order of integration can affect the outcome of the integral.

Why would someone use dxdy instead of dydx in a double integral?

The choice between dxdy and dydx in a double integral depends on the nature of the problem and the region being integrated over. In some cases, using dxdy may be more convenient or result in a simpler integral.

How do you find the area of a double integral using dxdy?

To find the area of a double integral using dxdy, you first need to set up the integral by identifying the limits of integration for both the x and y variables. Then, you evaluate the integral using standard integration techniques. The result will be the area of the region being integrated over.

Can dxdy and dydx be used interchangeably in a double integral?

No, dxdy and dydx cannot be used interchangeably in a double integral. The order of integration affects the outcome of the integral, so it is important to choose the appropriate one based on the problem at hand.

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