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June 30th, 2015, 03:26 AM   #1
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Double integral, repeated integral and the FTC

The fundamental theorem of calculus says that



And the green theorem says that



So, how to apply theses ideias theorem in this integral:



?

Last edited by Jhenrique; June 30th, 2015 at 03:46 AM.
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June 30th, 2015, 05:19 AM   #2
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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
Could you be more specific?
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June 30th, 2015, 05:47 AM   #3
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Notice that the integral along the area, A, is equal to the double integral along the two coordinates x and y, with dx and dy as integration variables, instead of dA.
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June 30th, 2015, 06:07 AM   #4
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I don't see that "Green's theorem" has anything to do with this. The way you do apply the "Fundamental Theorem" to double integrals is to separate the region of integration into "Type I" and "Type II" regions.

A "Type I" region is one in which there are "upper" and "lower" boundaries in which y is a function of x and the two boundaries do not cross. A "Type II" region is one in which there are "right" and "left" boundaries in which x is a function of y and the two boundaries do not cross.

For a "Type I" region, write those two boundaries as y= p(x) for the upper boundary and y= q(x) for the lower boundary. For that region, then, $\displaystyle \int\int f(x,y)dxdy= \int_a^b\left(\int_{q(x)}^{p(x)} f(x, y) dy\right) dx$ where a is the lowest value of x in that region and b is the largest.

To use the "fundamental theorem" now, find an "anti-derivative" for f(x, y) (as a function of x, treating y as a constant), F(x,y), and evaluate at p(x) and q(x): F(x, p(x))- F(x, q(x)). That is a function of x only which can now be integrated with respect to x in the "usual" way.

For a "Type II" region, write those two boundaries as x= p(y) for the right boundary and x= q(y) for the left boundary. For that region, then, $\displaystyle \int\int f(x, y)dxdy= \int_c^d\left(\int_{q(y)}^{p(y)} f(x, y) dx\right)dy$ where c is the lowest value of y in that region and q is the largest.

For example, to integrate a function, F(x,y), over a disc with center (0, 0) and radius R, we can think of that single region as either a "Type I" or a "Type II" region. As a type I region, x goes from -R to R and, for each x, $\displaystyle p(x)= \sqrt{R^2- x^2}$ and $\displaystyle q(x)= -\sqrt{R^2- x^2}$.
So the integral would be $\displaystyle \int_{-R}^R \int_{-\sqrt{R^2- x^2}}^{\sqrt{R^2- x^2}} F(x, y) dy dx$. Integrate F(x,y) as a function of y, treating x as a constant and evaluate at $\displaystyle \sqrt{R^2- x^2}$ and $\displaystyle -sqrt{R^2- x^2}$. That will result in an integral in x only.

As a "Type II" integral, write y goes from -R to R and, for each y, $\displaystyle p(y)= \sqrt{R^2+ y^2}$ and $\displaystyle q(x)= -\sqrt{R^2- y^2}$. The integral would be $\displaystyle \int_{-R}^{R}\int_{-\sqrt{R^2- y^2}}^{\sqrt{R^2- y^2}} F(x, y) dx dy$

In this particular example, the limits of integration are constants and the region of integration is a rectangle.
First integrate f(x,y) with respect to y only, treating x as a constant, to get the anti-derivative F(x,y). Evaluate that at y= d and y= c: F(x,d)- F(x,c). Now the integral is . Integrate to get an anti-derivative, G(x), of that. Evaluate between a and b: G(b)- G(a).

Last edited by Country Boy; June 30th, 2015 at 06:30 AM.
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June 30th, 2015, 06:27 AM   #5
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Actually, I did this question with this geometric problem in mind:



With S: f(x,y)

Is the analogous tridimensional of FTC.
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June 30th, 2015, 04:45 PM   #6
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Let see if now you understand...

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