Line integral of scalar field ( piecewise curve)

chetzread
Messages
798
Reaction score
1

Homework Statement



for the line segment c2 , why did the author want to differentiate dx with respect to dy ? and he gt dx = 0 ?

I'm curious why did he didnt do so for C3 , where dy= 0 ..Why didnt he also differentiate dy with dx ? dy/dx = 0 ?

Homework Equations

The Attempt at a Solution


is there anything wrong with the working ?
 

Attachments

  • 10.jpg
    10.jpg
    29.9 KB · Views: 404
  • 11.jpg
    11.jpg
    15 KB · Views: 401
Physics news on Phys.org
chetzread said:
for the line segment c2 , why did the author want to differentiate dx with respect to dy ? and he gt dx = 0 ?

actually one differentiates a function f(x,y) with respect to variable x or y;
your statement is in error - during c2 he is taking d/dy of f=x and as y is only variable on the path the slope of the path d/dy of x comes to zero.

chetzread said:
I'm curious why did he didnt do so for C3 , where dy= 0 ..Why didnt he also differentiate dy with dx ? dy/dx = 0 ?
here also the author is taking d/dx of the function f=y as the variable is x ...not y as in the previous path and the slope is again zero.
slope of a curve is defined by tan of the angle made by a drawn tangent at the point under consideration..
Imagine f equivalent to say F(x,y) ...the graph of f is of a type x=0 for the c2 ( y from 1 to zero) and y=0 for c3 ( x going from 0 to 1)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top