Physics homework help. I assume the following knowledge; please ask as separate question(s) if any of these are not already established:
Concept of partial derivatives
The area of a surface, ##f(x,y)##, above a region R of the XY-plane is given by ##int int_R sqrt((f_x’)^2 + (f_y’)^2 +1) dx dy## where
##f_x’## and ##f_y’## are the partial derivatives of ##f(x,y)## with respect to ##x## and ##y## respectively.
In converting the integral of a function in rectangular coordinates to a function in polar coordinates: ##dx dy rarr (r) dr d theta##
If ##z = f(x,y) = x^2 + y^2##
then ##f_x’ = 2x## and ##f_y’= 2y##
The Surface area over the Region defined by ##x^2+y^2 = 1##is given by
##S =int int_R sqrt(4x^2 + 4y^2 + 1) dx dy##
Converting this to polar coordinates (because it is easier to work with the circular Region using polar coordinates)
##S = int_(theta = 0)^(2pi) int_(r=0)^1 (4 r^2+1)^(1/2) (r) dr d theta##
##= int_(theta=0)^(2pi) ((4r^2+1)^(3/2))/(12) |_(r=0)^1 d theta##
##= int_(theta=0)^(2pi) (5sqrt(5)-1)/(12) d theta##
##= (5sqrt(5) -1)/(12) theta |_(theta=0)^(2pi)##
##= (5sqrt(5)-1)/6 pi##

Physics homework help

Physics homework help. I assume the following knowledge; please ask as separate question(s) if any of these are not already established:
Concept of partial derivatives
The area of a surface, ##f(x,y)##, above a region R of the XY-plane is given by ##int int_R sqrt((f_x’)^2 + (f_y’)^2 +1) dx dy## where
##f_x’## and ##f_y’## are the partial derivatives of ##f(x,y)## with respect to ##x## and ##y## respectively.
In converting the integral of a function in rectangular coordinates to a function in polar coordinates: ##dx dy rarr (r) dr d theta##
If ##z = f(x,y) = x^2 + y^2##
then ##f_x’ = 2x## and ##f_y’= 2y##
The Surface area over the Region defined by ##x^2+y^2 = 1##is given by
##S =int int_R sqrt(4x^2 + 4y^2 + 1) dx dy##
Converting this to polar coordinates (because it is easier to work with the circular Region using polar coordinates)
##S = int_(theta = 0)^(2pi) int_(r=0)^1 (4 r^2+1)^(1/2) (r) dr d theta##
##= int_(theta=0)^(2pi) ((4r^2+1)^(3/2))/(12) |_(r=0)^1 d theta##
##= int_(theta=0)^(2pi) (5sqrt(5)-1)/(12) d theta##
##= (5sqrt(5) -1)/(12) theta |_(theta=0)^(2pi)##
##= (5sqrt(5)-1)/6 pi##

Physics homework help