Shading the region implied by a double integration expression

[cooldudeachyut]

I can shade the region the expression ₀∫¹_x∫^√(2-x2)dydx finds "area" of, however cannot understand what region a similar expression implies where x varies from 0 to √2, i.e. ₀∫^√2_x∫^√(2-x2)dydx

So, what's the graphical significance of the later expression?

 

[Dr.Peterson]

I can shade the region the expression ₀∫¹_x∫^√(2-x2)dydx finds "area" of, however cannot understand what region a similar expression implies where x varies from 0 to √2, i.e. ₀∫^√2_x∫^√(2-x2)dydx

So, what's the graphical significance of the later expression?

It's the net signed area of the "shaded region" -- that is, the sum of a positive area and a negative "area".

 

[HallsofIvy]

The outer integral, with constant limits, 0 and 1, is with respect to x so mark x= 0 and x= 1 on the x-axis of an xy coordinate system. For each x then the inner integral is from y= x to \(\displaystyle y= \sqrt{2- x^2}\). y= x is easy. It is the line from (0, 0) to (1, 1) within the interval x= 0 to x= 1. Draw that line segment.

For \(\displaystyle y= \sqrt{2- x^2}\), square both sides to get \(\displaystyle y^2= 2- x^2\) or \(\displaystyle x^2+ y^2= 2\). You should recognize that as the equation of the circle with center at (0, 0) and radius \(\displaystyle \sqrt{2}\). Since a square root is always non-negative, the graph of \(\displaystyle y= \sqrt{2- x^2}\) is half of that- the semi-circle above the x-axis. When x= 0, \(\displaystyle y= \sqrt{2}\) and when x= 1, y= 1.

The region to be integrated is the region above the line y= x, x between 0 and 1, and below the portion of the circle from \(\displaystyle \left(0, \sqrt{2}\right)\) to \(\displaystyle \left(1, 1\right)\).