1. ## Double Integral Question

This example is from MIT OCW, Multivariable Calculus (http://ocw.mit.edu/courses/mathemati...C_notes_25.pdf)

Shouldn't the highlighted integral start from a, and not zero?

2. ## Re: Double Integral Question

In polar coordinates, when a point is represented as $(r, \theta)$, r is always the distance from the origin to that point. Here, integrating in polar coordinates, as $\theta$ goes from 0 to $\pi/4$, the corresponding r integral is from the origin to the point where the ray at angle $\theta$ crosses the boundary.

3. ## Re: Double Integral Question

At the origin (which is part of R), the distance to the origin is zero. The point on the edge of the square is a distance of $a\sec \theta$ from the origin. So, the integral should start from zero and go to $a\sec \theta$. As $\theta$ changes, it represents a different line segment of points in that triangle. The first line segment is from the origin along the x-axis to the edge of the square. The last one is the hypotenuse of the triangle.

4. ## Re: Double Integral Question

I got that. Thank you very much both of you.