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

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

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- Jul 16th 2014, 09:07 AMcosmonavtDouble Integral Question
This example is from MIT OCW, Multivariable Calculus (http://ocw.mit.edu/courses/mathemati...C_notes_25.pdf)

Attachment 31300

Shouldn't the highlighted integral start from a, and not zero? - Jul 16th 2014, 09:43 AMHallsofIvyRe: Double Integral Question
In polar coordinates, when a point is represented as $\displaystyle (r, \theta)$, r is

**always**the distance from the origin to that point. Here, integrating in polar coordinates, as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/4$, the corresponding r integral is from the origin to the point where the ray at angle $\displaystyle \theta$ crosses the boundary. - Jul 16th 2014, 09:49 AMSlipEternalRe: 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 $\displaystyle a\sec \theta$ from the origin. So, the integral should start from zero and go to $\displaystyle a\sec \theta$. As $\displaystyle \theta$ 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.

- Jul 16th 2014, 11:57 PMcosmonavtRe: Double Integral Question
I got that. Thank you very much both of you.