- #1

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How does this work? I can only think that this is some sort of u-substitution of r^2 for r but then why don't the interval values change to 0 and R^2? Thanks!

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- Thread starter sparkle123
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- #1

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How does this work? I can only think that this is some sort of u-substitution of r^2 for r but then why don't the interval values change to 0 and R^2? Thanks!

- #2

- 768

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Slight detail.

Also, don't let little r get confused with capital R. R is just some number.

- #3

SammyS

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I suppose the proper thing to do would have been to write the second line as in one of the following forms:

[itex]\displaystyle \int_0^{r=R}\left(x^2+r^2\right)^{-3/2}\,d(r^2)[/itex]

[itex]\displaystyle \int_0^{\sqrt{R}}\left(x^2+r^2\right)^{-3/2}\,d(r^2)\,,[/itex] since r^{2} is the "variable" of integration.

[itex]\displaystyle \int_0^{\sqrt{R}}\left(x^2+r^2\right)^{-3/2}\,d(r^2)\,,[/itex] since r

But it's not unusual to see it expressed the way you posted it.

- #4

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- #5

SammyS

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You can usually work out the indefinite integral using whatever methods, writing the final result in terms of the original variable, of course, then put in the bounds for the definite integral.

- #6

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opposite = r

adjacent = x

hypotenuse = sqrt(r^2 + x^2)

hypotenuse/opposite = csc(theta) = sqrt(r^2 + x^2)/r

csc(theta)^3 = (r^2 + x^2)^(3/2)/r^3

then

2 integral [0,R] r/(x^2+r^2)^(3/2) dr = 2 integral [0, R] dr/(r^3*csc(theta)^3)

I'll let you figure out were to go from there ^_^

- #7

HallsofIvy

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However, I would make the substitution for the entire root, not just r

let u= x

- #8

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Thanks everyone!

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