Examples of Double Integrals in Polar Coordinates (Example 2)

We want to find the value of the following definite integral by converting the double integral into polar coordinates.

\[\int_0^2 {\int\limits_0^{\sqrt {2 - {x^2}} } {\sqrt {{x^2} + {y^2}} dydx} } \]

See example 1, but the little difference here is the upper integral limit, it looks a little different. That's because the region of integration is a shifted circle, not just a circle that has a center in the origin.

An easy trick to knowing the geometry of the circle is to just look at the other boundaries, notice that the integral for x goes from 0 to 2. It'll always match with the region of the circle, otherwise the problem wouldn't make sense.

So you know that for this problem the region of integration is a circle that goes from x=0 to x=2, basically a semi-circle that's only in the first quadrant. In other words, notice that y goes from 0 to "some positive circle", and that x goes from 0 to 2. The only geometry that makes sense is a semicircle in the first quadrant.

We need to use substitution to convert this double integral to polar coordinates. The upper boundary for y can be used, we square that to get a relation for the radius. You can set that upper boundary, \(\sqrt {2 - {x^2}} \), to be equal to y. See the following steps:

\[\begin{array}{l}{x^2} + {y^2} = {r^2}\\x = r\cos (\theta )\\y = \sqrt {2x - {x^2}} \\{y^2} = 2x - {x^2}\\{x^2} + {y^2} = 2x\\{r^2} = 2x\\{r^2} = 2r\cos (\theta )\\r = 2\cos (\theta )\end{array}\]

So now that radius isn't really a simple answer, it's in terms of \(2\cos (\theta )\). It's because the circle was horizontally shifted, radius is only a clean number when the circle is centered at the origin. So if it's not immediately clear what the radius is, you may need to play around with a few trigonometric equations/relations.

So the radius goes from 0 to \(2\cos (\theta )\). The circle is only in the first quadrant, and goes from 0 to 2. So we need to include the entire first quadrant, meaning theta goes from 0 to \(\frac{\pi }{2}\). Don't forget to add an extra "r" when you change to polar coordinates, and when you do the substitution you obtain:

\[\int_0^{\frac{\pi }{2}} {\int_0^{2\cos (\theta )} {r \cdot rdrd\theta } } \]

And we work through it as usual. So do that inner integral first, and evaluate it at both boundaries. You get 1/3 r cubed, and you evaluate it at both boundaries, and then you need to do the final integral.

\[\int_0^{\frac{\pi }{2}} {\frac{8}{3}} {\cos ^3}\left( \theta \right)d\theta \]

It's an odd-powered cosine term. Break then apart, and use a relation to simplify. When you get a mixture or cosines and sines, then you can use substitution to easily solve.

\[\begin{array}{l}\int_0^{\frac{\pi }{2}} {\frac{8}{3}} {\cos ^2}\left( \theta \right)\cos \left( \theta \right)d\theta \\\frac{8}{3}\int_0^{\frac{\pi }{2}} {\left( {1 - {{\sin }^2}\left( \theta \right)} \right)} \cos \left( \theta \right)d\theta \end{array}\]

Use substitution and solve. Don't forget, the integral boundaries will change when you use substitution!

\[\begin{array}{l}u = \sin \left( \theta \right)\\du = \cos \left( \theta \right)d\theta \\\sin (0) = 0\\\sin \left( {\frac{\pi }{2}} \right) = 1\\\frac{8}{3}\int_0^1 {\left( {1 - {u^2}} \right)du} \end{array}\]

Then simplify and you will get a clean final answer:


Please see other examples of solving double integrals in polar coordinates below:

Double Integrals in Polar Coordinates (Ex1)

Double Integrals in Polar Coordinates (Ex3)

Double Integrals in Polar Coordinates (Ex4 Cylinders and Ellipsoids)

Double Integrals in Polar Coordinates (Ex5 Spheres and Cylinders)

Alternatively, return to the solving double integrals using polar coordinates hub page:

Double Integrals in Polar Coordinates

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