       RE: integral question

• To: mathgroup at smc.vnet.net
• Subject: [mg49853] RE: [mg49834] integral question
• From: "David Park" <djmp at earthlink.net>
• Date: Wed, 4 Aug 2004 10:46:31 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Doing this in some detail, we have for the original integral in terms of
r...

eqn0 = r^2 == x^2 + y^2
Sqrt /@ %
Simplify[%, r > 0]
eqn1 = Reverse[%]
giving
r^2 == x^2 + y^2
Sqrt[r^2] == Sqrt[x^2 + y^2]
Sqrt[x^2 + y^2] == r
r == Sqrt[x^2 + y^2]

Now finding the r limits along your line.

eqn1 /. Thread[{x, y} -> {0, 0}]
eqn1 /. Thread[{x, y} -> {1, 1}]
giving
r == 0
r == Sqrt

Integrate[r^2, {r, 0, Sqrt}]
(2*Sqrt)/3

Converting into an integral with x and dx. We first solve for dr (== Dt[r])

eqn0
% /. y -> x
Dt[%]
drrule = Solve[%, Dt[r]][[1,1]]
giving
r^2 == x^2 + y^2
r^2 == 2*x^2
2 r Dt[r] == 4 x Dt[x]
Dt[r] -> (2*x*Dt[x])/r

Then calculating the new integral in terms of x and dx...

r^2Dt[r]
% /. drrule
% /. r -> Sqrt[x^2 + y^2]
% /. y -> x
Simplify[%, x > 0]
giving
r^2*Dt[r]
2*r*x*Dt[x]
2*x*Sqrt[x^2 + y^2]*Dt[x]
2*Sqrt*x*Sqrt[x^2]*Dt[x]
2*Sqrt*x^2*Dt[x]

We can then do the integral over x.

Integrate[2*Sqrt*x^2, {x, 0, 1}]
(2*Sqrt)/3

David Park

From: quest04 [mailto:na at na.na]
To: mathgroup at smc.vnet.net

Hello,
I have a question and was wondering if someone could help me with it.  This
is a general vector calculus problem, not specific to Mathematica.  I have a
simple integral as follows:
Given r^2= x^2+y^2, solve Integral[r^2, dr] from point1 (0,0) to point2
(1,1), which would be evaluated from r=0 to r=sqrt and gives answer =
2*sqrt/3.
The above is pretty simple, however, I am not sure how to formulate the
problem when I convert the 'dr' back to cartesian coordinates as follows:
Integrate [x^2+y^2, d????] and the limits???  WHat should 'dr' be in terms
of dx? if my integrand is directly x^2+y^2.  It cannot be a double integral,
since the original problem was a single integral, but then how do you solve
this??
Can anyone help?

thanks!

```

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