So I was just asked this question on an exam and I have no idea how to convert the equation into polar form.
Question was something along the lines of:
Integrate f(x,y)=1/x over the curve r=2cos(theta) 0 <= theta <= pi/2
So I know that this is a double integral and I should be integrating 1/x over the domain defined by r, times the differential of area with respect to x and y converted to polar differentials. If I'm not mistaken, f(x,y)=1/x is a three dimensional surface extending infinitely in the y direction and following the shape of z=1/x in the x-z plane.
I tried substituting rcos(theta) for f(x,y) giving 1/(r cos(theta)) but it is not integrable over 0 <= theta <= pi/2 so clearly wasn't the right approach.
Question was something along the lines of:
Integrate f(x,y)=1/x over the curve r=2cos(theta) 0 <= theta <= pi/2
So I know that this is a double integral and I should be integrating 1/x over the domain defined by r, times the differential of area with respect to x and y converted to polar differentials. If I'm not mistaken, f(x,y)=1/x is a three dimensional surface extending infinitely in the y direction and following the shape of z=1/x in the x-z plane.
I tried substituting rcos(theta) for f(x,y) giving 1/(r cos(theta)) but it is not integrable over 0 <= theta <= pi/2 so clearly wasn't the right approach.