In the last section, we learned how to compute triple integrals. We also saw that integrals over circular regions, like in Example 4 (the two paraboloids), are extremely difficult to compute in rectangular $(x,y,z)$ coordinates.
Today, we introduce our first new 3D coordinate system designed to solve this exact problem. It's called Cylindrical Coordinates, and the main idea is simple: it's just polar coordinates plus a $z$-axis.
In 2D, we describe a point $P$ by its rectangular coordinates $(x,y)$, which represent horizontal and vertical distance.
Polar coordinates describe the same point $P$ using $(r, \theta)$, where:
Two perspectives on the relationship between rectangular $(x,y)$ and polar $(r,\theta)$ coordinates.
Based on the right-triangle trigonometry shown in the graph, we get our conversion formulas:
From Polar to Rectangular:
$$ x = r \cos(\theta) \quad \text{and} \quad y = r \sin(\theta) $$
From Rectangular to Polar:
$$ r^2 = x^2 + y^2 \quad \text{and} \quad \tan(\theta) = \frac{y}{x} $$
When we change variables in a double integral, we must also change the area element $dA = dx \,dy$.
A common mistake is to assume $dA = dr \,d\theta$. This is incorrect. A small "rectangle" in polar coordinates is not a true rectangle—it's a small, curved sector.
A "polar rectangle" with width $\Delta r$ and arclengths $r\Delta\theta$ and $(r+\Delta r)\Delta\theta$.
This shape has two straight sides of length $\Delta r$. Its other two sides are arcs. The inner arc has length $r \Delta\theta$, and the outer arc has length $(r+\Delta r)\Delta\theta$.
For a very small element, we can approximate its area as a rectangle with width $\Delta r$ and length $r \Delta\theta$.
$$ \Delta A \approx (r \, \Delta\theta) \cdot (\Delta r) = r \, \Delta r \, \Delta\theta $$
The area element $dA$ in polar coordinates is:
$$ dA = \boldsymbol{r} \,dr \,d\theta $$
You must include this extra $\boldsymbol{r}$. Think of it as a "toll" or a "conversion fee" you have to pay for switching from rectangular to polar. This is the most common mistake on exams!
To convert a double integral $\iint_D f(x,y) \,dA$ to polar coordinates, we do three things:
$$ \iint_D f(x,y) \,dA = \int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta) \boldsymbol{r} \,dr \,d\theta $$
In our last lecture (Example 4), we needed to solve $V = \int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (8 - 2(x^2+y^2)) \,dy \,dx$. Let's solve it now.
1. Convert the function: $f(x,y) = 8 - 2(x^2+y^2)$. Since $r^2 = x^2+y^2$, this becomes $f(r, \theta) = 8 - 2r^2$.
2. Convert the differential: $dy \,dx$ becomes $r \,dr \,d\theta$.
3. Convert the region $D$: The bounds $-2 \le x \le 2$ and $-\sqrt{4-x^2} \le y \le \sqrt{4-x^2}$ describe a circle of radius 2, $x^2+y^2 \le 4$. In polar, this region is described by constant bounds: $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.
The new integral is:
$$ V = \int_0^{2\pi} \int_0^2 (8 - 2r^2) \cdot \boldsymbol{r} \,dr \,d\theta $$
$$ V = \int_0^{2\pi} \int_0^2 (8r - 2r^3) \,dr \,d\theta $$
Inner ($dr$): $\int_0^2 (8r - 2r^3) \,dr = \left[ 4r^2 - \frac{r^4}{2} \right]_0^2 = (16 - \frac{16}{2}) - 0 = 8$.
Outer ($d\theta$): $\int_0^{2\pi} 8 \,d\theta = [8\theta]_0^{2\pi} = 16\pi$.
Set up, but do not evaluate, a polar integral for the area of the region $D$ enclosed by the cardioid $r = 1 + \cos(\theta)$.
Cylindrical coordinates are just polar coordinates with a $z$-axis. They are incredibly simple and intuitive.
A point $P(x,y,z)$ is described by $(r, \theta, z)$ where:
Cylindrical coordinates $(r, \theta, z)$. Move the sliders for R, T, and C to see how they determine the point's location.
From Cylindrical to Rectangular:
$$ x = r \cos(\theta) \quad \quad y = r \sin(\theta) \quad \quad z = z $$
From Rectangular to Cylindrical:
$$ r^2 = x^2 + y^2 \quad \quad \tan(\theta) = \frac{y}{x} \quad \quad z = z $$
Notice that $z$ doesn't change!
This is the most important part. To find the volume element $dV$, we just extend our 2D polar area element $dA$ upwards by a small height $dz$.
A "cylindrical box" is a "polar rectangle" with a small thickness $dz$.
$$ \Delta V = (\text{Base Area}) \cdot (\text{Height}) \approx (r \, \Delta r \, \Delta\theta) \cdot (\Delta z) $$
The cylindrical volume element $dV$ has a base $dA = r\,dr\,d\theta$ and height $dz$.
The volume element $dV$ in cylindrical coordinates is:
$$ dV = \boldsymbol{r} \,dz \,dr \,d\theta $$
Just like in polar, you must include the extra $\boldsymbol{r}$. It comes from the polar area of the "floor" of the volume element.
We use cylindrical coordinates when the solid $E$ (or its shadow $D$) has circular symmetry (e.g., cylinders, cones, paraboloids, spheres).
We almost always set up the integral in the order $\boldsymbol{r \,dz \,dr \,d\theta}$.
Evaluate the integral $\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{0}^{x^2+y^2} (x^2+y^2) \,dz \,dy \,dx$.
This integral is nearly impossible in rectangular coordinates. But let's look for our "triggers." The function is $f(x,y,z) = x^2+y^2$. The $z$-bounds go up to $z = x^2+y^2$. The $y$-bounds are $y = \pm\sqrt{4-x^2}$. All of these are "flashing red lights" to convert to cylindrical.
1. $z$-limits: The floor is $z=0$. The ceiling is $z = x^2+y^2 \implies z=r^2$. So, $0 \le z \le r^2$.
2. Shadow $D$: The $x$ and $y$ bounds, $\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \dots \,dy \,dx$, describe a circle of radius 2 centered at the origin.
3. $r, \theta$ limits: The shadow $D$ is $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.
4. Convert $f$ and $dV$: The function is $f = x^2+y^2 \implies f = r^2$. The volume element $dV$ becomes $r \,dz \,dr \,d\theta$.
Set up and Integrate:
$$ V = \int_0^{2\pi} \int_0^2 \int_0^{r^2} (r^2) \cdot \boldsymbol{r} \,dz \,dr \,d\theta $$
$$ V = \int_0^{2\pi} \int_0^2 \int_0^{r^2} r^3 \,dz \,dr \,d\theta $$
Inner ($dz$): $\int_0^{r^2} r^3 \,dz = r^3 [z]_0^{r^2} = r^3(r^2) = r^5$.
Middle ($dr$): $\int_0^2 r^5 \,dr = \left[ \frac{r^6}{6} \right]_0^2 = \frac{64}{6} = \frac{32}{3}$.
Outer ($d\theta$): $\int_0^{2\pi} \frac{32}{3} \,d\theta = \left[\frac{32}{3}\theta\right]_0^{2\pi} = \frac{64\pi}{3}$.
What was impossible in rectangular becomes a straightforward computation in cylindrical.
The solid $E$ bounded by the paraboloid $z = x^2+y^2$ and the plane $z=4$ .
Find the volume of the solid $E$ bounded by the paraboloid $z = x^2+y^2$ and the plane $z=4$. (This was CYU #2 from our last lecture).
1. $z$-limits: The "floor" is $z_1 = x^2+y^2$. The "ceiling" is $z_2 = 4$. In cylindrical, $z_1 = r^2$. So, $r^2 \le z \le 4$.
2. Shadow $D$: The shadow is the intersection of the two surfaces: $x^2+y^2 = 4$. This is a circle of radius 2.
3. $r, \theta$ limits: The region $D$ is $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.
4. Convert $f$ and $dV$: We are finding volume, so $f(x,y,z) = 1$. $dV$ becomes $r \,dz \,dr \,d\theta$.
Set up and Integrate:
$$ V = \int_0^{2\pi} \int_0^2 \int_{r^2}^4 1 \cdot \boldsymbol{r} \,dz \,dr \,d\theta $$
Inner ($dz$): $\int_{r^2}^4 r \,dz = r [z]_{r^2}^4 = r(4 - r^2) = 4r - r^3$.
Middle ($dr$): $\int_0^2 (4r - r^3) \,dr = \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 = (2(4) - \frac{16}{4}) - 0 = 8 - 4 = 4$.
Outer ($d\theta$): $\int_0^{2\pi} 4 \,d\theta = [4\theta]_0^{2\pi} = 8\pi$.
Find the mass of the solid cone $E$ bounded by $z = \sqrt{x^2+y^2}$ and $z=1$, if the density is $\rho(x,y,z) = z$.
1. $z$-limits: The "floor" is the cone $z = \sqrt{x^2+y^2}$, which in cylindrical is $z = r$. The "ceiling" is the plane $z = 1$. So, $r \le z \le 1$.
2. Shadow $D$: The shadow is the intersection: $r = 1$. This is a circle in the $xy$-plane of radius 1.
3. $r, \theta$ limits: The region $D$ is $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.
4. Convert $f$ and $dV$: The function is $\rho = z$. $dV$ becomes $r \,dz \,dr \,d\theta$.
Set up and Integrate:
$$ m = \int_0^{2\pi} \int_0^1 \int_r^1 z \cdot \boldsymbol{r} \,dz \,dr \,d\theta $$
Inner ($dz$): $\int_r^1 rz \,dz = r \left[ \frac{z^2}{2} \right]_r^1 = r \left( \frac{1}{2} - \frac{r^2}{2} \right) = \frac{1}{2}(r - r^3)$.
Middle ($dr$): $\int_0^1 \frac{1}{2}(r - r^3) \,dr = \frac{1}{2} \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_0^1 = \frac{1}{2} \left( (\frac{1}{2} - \frac{1}{4}) - 0 \right) = \frac{1}{2} \left( \frac{1}{4} \right) = \frac{1}{8}$.
Outer ($d\theta$): $\int_0^{2\pi} \frac{1}{8} \,d\theta = \left[\frac{1}{8}\theta\right]_0^{2\pi} = \frac{2\pi}{8} = \frac{\pi}{4}$.
Set up an integral for the volume of the "ice cream cone" $E$ cut from the solid sphere $x^2+y^2+z^2 \le 4$ by the cone $z = \sqrt{x^2+y^2}$.
The solid $E$ is bounded below by the cone (red) and above by the sphere (blue).
Today we learned how to use cylindrical coordinates. This is our first major tool for simplifying triple integrals. We saw that it is just an extension of polar coordinates, and the key is to correctly identify the bounds and to always include the $r$ in the volume element $dV = r \,dz \,dr \,d\theta$.
After this lecture, you should be able to:
Test your knowledge by trying to solve these problems.
Evaluate $\iiint_E x \,dV$ where $E$ is the solid region in the first octant bounded by $z=0$, $z=x^2+y^2$, and the cylinder $x^2+y^2=1$.
Set up the integral for the volume of the solid bounded by $z=0$ and the paraboloid $z = 9 - x^2 - y^2$.
Set up the integral for the mass of a solid cylinder $E$ defined by $x^2+y^2 \le 4$, $0 \le z \le 3$, with density $\rho(x,y,z) = x^2+y^2+z^2$.