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We are currently interested in computing integrals of functions over various
regions in and via Some regions like rectangles and boxes are easy to
describe using -coordinates (a.k.a. rectangular coordinates). However,
other regions like circles and other things with rotational symmetry are
easier to work with in polar coordinates. Recall that in polar coordinates,
where is a function of . When working with parametric equations of this form, it is
common to notate and state that we are working in polar coordinates.
An ordered pair consisting of a radius and an angle can be graphed as
Coordinates of this type are called polar coordinates.
Consider the point in polar coordinates. What is this point when expressed in
Consider the point in -coordinates. What is this point when expressed in polar
coordinates with ?
Double integrals in polar coordinates
The basic form of the double integral is:
which can be interpreted as
Over some region, sum up products of heights and areas.
Of course if you want to evaluate the integral (and honestly, who doesn’t?) you have
to change to a region defined in -coordinates, and change to or leaving iterated
Now consider representing a region with polar coordinates.
Let be the region in the first quadrant bounded by the curve. We can approximate
this region using the natural shape of polar coordinates: Portions of sectors of circles.
In the figure, one such region is shaded, shown below:
From the picture above, we see that:
Recalling that the determinant of a matrix gives the area of a parallelogram, we could
also deduce the correct formula for by setting
So to evaluate replace with and convert the function to a function of polar
coordinates: Finally, find bounds and that describe . Let’s state this as a
Fubini Let be continuous on the region Then:
Write down a double integral in polar coordinates that will compute the area of a
circle of radius .
When first exploring problems like
this, we should try to plot them if we can.
Now either by reasoning about cosine, or by experimenting with the
plot, we should convince ourselves that if , the area is given by:
Finally, let’s derive the volume of a sphere using a double integral in polar
Find the volume of a sphere with radius .
The sphere of radius , centered at the
origin, has equation Solving for , we have . This gives the upper half of a sphere. We
wish to find the volume under this top half, then double it to find the total
The region we need to integrate over is the circle of radius , centered
at the origin. Thus, the volume of a sphere with radius is:
The formula for the volume of a sphere with radius is given as . We have justified
this formula with our calculation!
One may wonder how polar coordinates could be extended to triple integrals…read