Home»Math Guides»Double Integrals using Polar Coordinates (sphere and cylinder region of integration)
Examples of Double Integrals in Polar Coordinates in Sphere and Cylinder regions of integration
We can solve integrals and use polar coordinates to find the volume inside of multiple geometry shapes, such as a volume contained in one sphere, but outside a cylinder.
Example Problem: Find the volume inside the sphere \({x^2} + {y^2} + {z^2} = 16\) and outside the cylinder \({x^2} + {y^2} = 4\)
We can just use polar coordinates to do this, even though it’s a 3-dimensional problem!
We just need to find a way to convert these x,y, and z coordinates into radius and theta coordinates.
The rule is to take the equation that has “z”, isolate for “z” and put it in the integral, and use the boundaries from the other equation.
We start with the sphere equation and isolate it in terms of “z”.
The sphere is symmetric with respect to the “z” axis.
When you attempt to isolate it in terms of “z”, you get a plus or minus symbol, but you can just take it as a positive and multiply it by two to get the volume since it’s symmetric.
And we can substitute in polar coordinates to get x and y in terms of “r”.
\[\begin{array}{l}{x^2} + {y^2} = 4\\{x^2} + {y^2} + {z^2} = 16\\z = \pm \sqrt {16 – {x^2} – {y^2}} \\z = 2\sqrt {16 – {r^2}} \end{array}\]
This is what we will put in the double integral.
We need the volume inside the sphere, but outside the cylinder.
The constants in each equation represents the radius squared, so take the square root of to get the radii we need to use.
The root of 16 is 4, and the root of 4 is 2. So radius is between 2 and 4. Radius cannot be a negative value.
There are no restrictions on the quadrant/octant used, so theta is from 0 to \(2\pi \).
We need to multiply with an extra “r” term when converting to polar coordinates.
We end up with the following double integral (don’t forget we multiplied by a factor of two and we can take it out since it’s a constant). Then use regular substitution to solve.
\[V = 2\int_0^{2\pi } {\int\limits_2^4 {\left( {\sqrt {16 – {r^2}} } \right)rdrd\theta } } \]
\[\begin{array}{l}u = 16 – {r^2}\\du = – 2rdr\\u(2) = 16 – 4 = 12\\u(4) = 16 – 16 = 0\\V = – 2\int_0^{2\pi } {\int\limits_{12}^0 {\left( {\frac{1}{2}{u^{\frac{1}{2}}}} \right)dud\theta } } \\V = – 2\frac{1}{2}\int_0^{2\pi } {\frac{2}{3}} {(0)^{\frac{3}{2}}} – \frac{2}{3}{(12)^{\frac{3}{2}}}d\theta \\V = – 2 \cdot \frac{1}{2} \cdot \frac{2}{3}{\int_0^{2\pi } {0 – \left( {12} \right)} ^{\frac{3}{2}}}d\theta \\V = – 2 \cdot \frac{2}{3} \cdot – {\left( {12} \right)^{\frac{3}{2}}} \cdot 2\pi \cdot \frac{1}{2}\\V = \frac{{4\pi {{\left( {12} \right)}^{\frac{3}{2}}}}}{3}\\V = \frac{{96\sqrt 3 }}{3}\\V = 32\sqrt 3 \end{array}\]
For the last few steps, you can simplify the volume by using knowledge of how powers work.
For example, you have root 12 to the power of three, so two of those will be 12.
As well, factor out a four from the square root and take it out by square rooting it as 2.
This is a good example of a math problem where the final answer looks messy at first, but can actually be simplified.
If you’ve done integration with all kinds of shapes, try line integral examples next!
If you’re comfortable with integration, try examples with multi-variable partial derivatives next!