Homework Help: Stuck on the integration bound

How did you get those integration limits of ##[\sqrt 2,2]## for r? ##r## is the distance from the origin to:

• the line ##y=1## for the lower limit and
• the circle boundary for the upper limit

Looking at the diagram, we see that, at ##\theta=0## the integration is for ##r\in[1,2]## and for ##\theta=\pi/4## it is for ##r\in [\sqrt 2,\sqrt 2]##. Those limits match the limits in the solution formula, but not the ones in the integral above it.

 

Draw a line segment from the origin, at an angle of ##\theta## above the x axis, to the point where it hits the curve.
The lower limit for the inner integral is the distance along that segment from the origin (call it O) to the point where the segment hits the vertical line ##x=1## (call that point Q).
The upper limit for the inner integral is the distance from O to where the line segment hits the curve.
Use geometry to calculate those two lengths as formulas in terms of ##\theta##.

 

Let the shaded shape in your diagram be ABCA, where AB and CA are horizontal and vertical line segments and BC is the circular arc. The origin is O.

One way to solve it using polar coordinates is to evaluate the integral over OBCO in polar coordinates, then subtract the integral over OACO.