# Double integrals



My notes of the excellent lectures 16, 17 and 18 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Recall: the integral of function of one variable $$f(x)$$ corresponds to the area below the graph of $$f$$ over $$[a,b]$$.

$$\int_a^b f(x)\,dx \nonumber$$

The input domain of $$f(x)$$ is $$x$$, therefore the region of integration $$R$$ is on a line along the $$x$$-axis. Here $$x=a$$ is the lower bound, and $$x=b$$ is the upper bound.

## Definition

For a function of two variables $$f(x,y)$$, the region of integration $$R$$ is bounded by a curve on the $$xy$$-plane. Using a double integral, you can find the volume between the region and a function $$z=f(x,y)$$.

To compute the volume, start with cutting the area of $$R$$ in small pieces $$\Delta A=\Delta y\Delta x$$

Consider all the pieces, and take the limit $$\Delta A_i\to 0$$. $$\lim_{\Delta A_i\to 0}\sum_i f(x_i,y_i)\,\Delta A_i \nonumber$$

Let $$dA=dy\,dx$$ be a tiny piece of area in region $$R$$. This gives the definition of the double integral of $$f(x,y)$$ over region $$R$$. $$\shaded{ \iint_R f(x,y)\,dA } \nonumber$$

Double integrals are evaluated as two embedded integrals, starting with the inner integral $$\int_{x_{min}}^{x_{max}} \underbrace{ \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy }_{\text{function of only }y} \,dx \nonumber$$ The bound functions encode the shape of region $$R$$.

The bounds of the inner integral might be functions of the outer variables.

## In Cartesian coordinates

To compute $$\iint_R f(x,y)\,dA$$, we take slices that scan the volume from the back to the front.

For the outer integral, let $$S(x_i)$$ be the area of a slice $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum\ yz$$-plane (the area of the thin purple vertical wall in the picture on the left). Then, the volume of each slice is $$S(x_i)\,\Delta x$$. The total volume follows as \begin{align} \rm{volume} &= \lim_{\Delta x\to 0}\sum_i S(x)\,\Delta x \nonumber \\ &=\int_{x_{min}}^{x_{max}} \underline{S(x)}\,dx \label{eq:doublecomp1} \end{align}

For the inner integral, $$x$$ is constant and $$y$$ is the variable of integration. For the range of $$y$$, we go from the far left to the far right on the given slice, as shown in the picture on the right $$S(x) = \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy \label{eq:doublecomp2}$$ Note that these inner bounds depend on $$x$$.

Substituting equation $$\eqref{eq:doublecomp2}$$ in $$\eqref{eq:doublecomp1}$$ give the iterated integral $$\shaded{ \iint_R f(x,y)\,dA = \int_{x_{min}}^{x_{max}} \left[ \int_{y_{min}(x)}^{y_{max}(x)} f(x,y)\,dy \right] dx } \nonumber$$

### Examples

#### One

Integrate $$z=1-x^2-y^2$$ over the region \left\{\begin{align*} 0\leq &x\leq 1 \\ 0\leq &y\leq 1 \end{align*}\right. \nonumber

Plot

The bounds are trivial \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\int_0^1 1-x^2-y^2\,dy}\,dx \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^1 1-x^2-y^2\,dy &= \left[ y-x^2y-\frac{y^3}{3} \right]_{y=0}^1 \\ &= (1-x^2-\frac{1}{3}) – 0 \\ &= \underline{\frac{2}{3}-x^2} \end{align*} \nonumber

Substituted back in the outer integral \begin{align*} \iint_R z(x,y)\,dA &=\int_0^1 \underline{\frac{2}{3}-x^2}\,dx \\ &=\left[\frac{2}{3}x-\frac{x^3}{3}\right]_{x=0}^1 = \frac{1}{3} \end{align*} \nonumber

#### Two

Integrate $$z=1-x^2-y^2$$ over the quarter unit disk region \left\{\begin{align*} x^2 + y^2 &\leq 1 \\ x &\geq 0 \\ y &\geq 0 \end{align*}\right. \nonumber

Plot

Find the bounds of integration

1. For $$\int dy$$, the inner integral, express the bounds of $$y$$ as a function of $$x$$. The lower bond is $$0$$. The upper bounds are on a quarter circle with $$x^2+y^2 = 1 \Rightarrow y=\sqrt{1-x^2}$$.
2. For $$\int dx$$, the outer integral, the range for $$x$$ is $$0$$ to $$1$$.

Fill in the bounds of the integrals \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\int_0^{\sqrt{1-x^2}} 1-x^2-y^2\,dy}\,dx \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^{\sqrt{1-x^2}} 1-x^2-y^2\,dy &= \left[ y-x^2y-\frac{y^3}{3} \right]_{y=0}^{\sqrt{1-x^2}} \\ &= \left(\sqrt{1-x^2}-x^2\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{3/2}\right) – 0 \\ &= (1-x^2)(1-x^2)^{1/2}-\frac{1}{3}(1-x^2)^{3/2} \\ &= \underline{\frac{2}{3}\left(1-x^2 \right)^{3/2}} \end{align*} \nonumber

Substitute back in the outer integral \begin{align*} \iint_R z(x,y)\,dA &= \int_0^1\underline{\frac{2}{3}\left(1-x^2 \right)^{3/2}}\,dx \end{align*} \nonumber

For computing the outer integral, substitute $$x=\sin\theta$$ and using the double angle formula $$cos^2\theta=\frac{1}{2}(1+\cos2\theta)$$ twice. This will eventually lead to the answer $$\frac{\pi}{8}$$.

As we will see later, using polar coordinates will be much easier!

## Changing the order of integration

We change the order of integration, when it makes it easier to compute the double integral.

### Examples

#### One

When the bounds are numbers, they form a rectangle and we can simply switch the order of integration $$\int_0^1\int_0^2 dx\,dy = \int_0^2\int_0^1 dy\,dx \nonumber$$

#### Two

The written way can’t be computed. Change the order of integration. $$\int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx \nonumber$$

Plot the region based on the existing bounds.

For the new inner integral, $$y$$ is constant and $$x$$ is the variable of integration. The old upper bound $$y=\sqrt{x} \Rightarrow x=y^2$$, and lower bound $$y=x \Rightarrow x=y$$ \begin{align*} \int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx &= \int_0^1 \underline{\int_{y^2}^y \frac{e^y}{y}dx}\,dy \\ \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_{y^2}^y \frac{e^y}{y}dx &= \left[x\frac{e^y}{y}\right]_{x=y^2}^y \\ &=e^y – e^y y \end{align*} \nonumber

Find the antiderivative for $$e^y – e^y y$$ (or use integrating by parts) \begin{align*} \left(y\,e^y\right)’ &= 1.e^y+y.(e^y)’=e^y+y\,e^y \\ \Rightarrow \left(-y\,e^y\right)’ &= -e^y-y\,e^y \\ \Rightarrow \left(-y\,e^y+2\,e^y\right)’ &= -e^y-y\,e^y + 2e^y \\ &= e^y-y\,e^y \end{align*} \nonumber

The outer integral evaluates to \begin{align*} \int_0^1\int_x^{\sqrt{x}} \frac{e^y}{y}dy\,dx &= \int_0^1 (e^y – e^y y)\,dx \\ &=\Big[ -y\,e^y + 2\,e^y \Big]_{y=0}^1 \\ &= (-1.e^1+2e^1)-(0+2.e^0) \\ &= e -2 \end{align*} \nonumber

#### Three

Exchange the order of integration to $$dx\,dy$$ for $$\int_0^1\int_x^{2x}f\,dy\,dx \nonumber$$

Plot the region based on the existing bounds.

These not simply connected regions results in two terms: $$0\lt y\lt 1$$ and $$1\lt y\lt 2$$. Each with different bounds for $$x$$ $$\int_0^1\int_x^{2x}f\,dy\,dx = \int_{0}^{1}\int_{y/2}^{y} f\,dx\,dy + \int_{1}^{2}\int_{y/2}^{1} f\,dx\,dy \nonumber$$

## In polar coordinates

In general, you switch to polar coordinates because the region is easier to setup, or the integrand becomes simpler.

Polar coordinates express point $$(x,y)$$ in the plane, using $$r$$ for the distance from the origin $$r$$, and $$\theta$$ as the counterclockwise angle with the positive $$x$$-axis. \shaded{ \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \end{align*} } \nonumber

### Area element

The area element $$\Delta A$$ is almost rectangular as shown below

One side is $$\Delta r$$ and the other side is $$r\,\Delta\theta$$. For the limit where $$\Delta\theta,r\to 0$$, the area element becomes $$\shaded{ dA=r\,dr\,d\theta } \nonumber$$

The double integral in polar coordinates $$\shaded{ \int_{\theta_{min}}^{\theta_{max}} \int_{r_{min}}^{r_{max}} f(r,\theta)\,r\,dr\,d\theta } \nonumber$$

### Examples

#### One

Redo the earlier problem using polar coordinates: Integrate $$z=1-x^2-y^2$$ over the quarter unit disk region \left\{\begin{align*} x^2 + y^2 &\leq 1 \\ x &\geq 0 \\ y &\geq 0 \end{align*}\right. \nonumber

Plot of the region

Set the bounds for the integrals

1. For $$\int dr$$, the inner integral: fix the value of $$\theta$$, and let $$r$$ vary. For the bounds, ask yourself for what values of $$r$$ will I be inside my region. In this case, that is $$0\lt r\lt 1$$. We let $$\theta$$ vary.
2. For $$\int d\theta$$, the outer integral: ask yourself what values of $$\theta$$ will I be inside my region.

Fill in the bounds of the double integral $$\int_0^{\pi/2}\int_0^1 f(r,\theta)\,r\,dr\,d\theta \nonumber$$

Instead of just replacing $$x=r\,\cos\theta$$ and $$y=r\,\sin\theta$$, we can express the function $$f(x,y)$$ in polar coordinates using $$r^2=x^2+y^2$$ \begin{align*} f(x,y) &= 1-x^2-y^2 \\ &= 1-(x^2+y^2) \\ \Leftrightarrow f(r,\theta) &= 1-r^2 \end{align*} \nonumber

Evaluate the double integral \begin{align*} \text{volume} &= \int_0^{\pi/2}\underline{\int_0^1 (1-r^2)\,r\,dr}\,d\theta \\ &= \int_0^{\pi/2}\left[ \frac{r^2}{2}-\frac{r^4}{4} \right]_{r=0}^1 \,d\theta \\ &= \int_0^{\pi/2} \frac{1}{4} \,d\theta = \frac{1}{4}\frac{\pi}{2}=\frac{\pi}{8} \end{align*} \nonumber

## Applications

### Find the area

Find the area of region $$R$$. $$\shaded{ \text{Area}(R)=\iint_R 1\,dA } \nonumber$$

Or, the mass of a (flat) object with density $$\delta$$ = mass per unit area. \shaded{ \begin{align*} \Delta m &= \delta .\Delta A \\ \Rightarrow \text{Mass}(R) &= \iint_R\delta(x,y)\,dA \end{align*} } \nonumber

### Find the average value

Average value of $$f$$ in $$R$$. $$\shaded{ \bar f = \frac{1}{\text{Area}(R)}\iint_R f(x,y)\,dA } \nonumber$$

Or, the weighted average value of $$f$$ in $$R$$ with density $$\delta$$ $$\shaded{ \frac{1}{\text{Mass}(R)}\iint_R f(x,y)\,\underbrace{\delta(x,y)\,dA}_{\text{mass element}} } \nonumber$$

Or, the center of mass $$(\bar x,\bar y)$$ of a (planar) object with density $$\delta$$. The weighted averages on $$x$$ and $$y$$ \shaded{ \left\{ \begin{align*} \bar x &= \iint_R x\,\delta(x,y)\,dA \\ \bar y &= \iint_R y\,\delta(x,y)\,dA \end{align*} \right. } \nonumber

### Find the moment of inertia

Recall from physics:

The kinetic energy of a point mass equals $$\frac{1}{2}mv^2$$

Mass is how hard it is to impart a translation movement. (to make it move)

Similarly, the moment of inertia about an axis is how hard it is to rotate about that axis (to make it spin).

Let $$\omega$$ be the rate of change of angle $$\theta$$, $$\omega=\frac{d\theta}{dt}$$.

At unit time, a mass $$m$$ rotating by $$\omega$$, goes a distance of $$r\omega$$, so the speed is $$v=r\omega$$. The kinetic energy follows as $$\shaded{ \tfrac{1}{2}m\,v^2=\tfrac{1}{2}\underline{mr^2}\omega^2 } \nonumber$$

The moment of inertia is defined as $$\shaded{ I = mr^2 } \nonumber$$

For rotation movements, $$I$$ replaces the mass $$m$$. The rotational kinetic energy is $$\shaded{ \frac{1}{2}\,I\,\omega^2 } \nonumber$$

A solid with density $$\delta_i$$ rotating about the origin.

A tiny area $$\Delta A$$ with mass $$\Delta m=\delta_i\,\Delta A$$, has a moment of inertia $$\Delta m.r^2=\delta.\Delta A.r^2 \nonumber$$

Consider all the pieces $$\shaded{ I_o=\iint_R r^2\,\delta\,dA } \nonumber$$ where $$r^2=x^2+y^2$$ in $$xy$$-coordinates.

#### Rotation about the $$\ x$$-axis

In the $$xyz$$-space, the distance to the $$x$$-axis is $$|y|$$.

Moment of inertia for a solid with density $$\delta$$ rotaring about the $$x$$-axis $$\shaded{ I_x=\iint_R y^\,\delta\,dA } \nonumber$$

#### Examples

##### One

Disk of radius $$a$$ with uniform density $$\delta=1$$ spinning around its center. What is the moment of inertia?

What is $$r^2$$ for any point inside $$R$$ in this formula? $$I_o=\iint_R r^2\,\delta\,dA \nonumber$$

Using polar coordinates, $$r$$ will go from $$0$$ to $$a$$ and $$dA=r\,dr\,d\theta$$ \begin{align*} I_o &= \iint_R r^2.1.dA \\ &= \int_0^{2\pi} \underline{\int_a^a r^2 r\,dr}\,d\theta \\ &= \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_{r=0}^a\,d\theta = \int_0^{2\pi} \frac{a^4}{4}\,d\theta \\ &= \frac{a^2}{4}\Big[\theta\Big]_0^{2\pi} = \frac{1}{2}\pi a^4 \end{align*} \nonumber

##### Two

How much harder is it to spin this disk around a point on its circumference?

The inertia \begin{align*} I_o & =\iint r^2\,dA \\ &= \int_{-\pi/2}^{\pi/2} \underline{\int_0^{2a\cos\theta} r^2 r\,dr}\,d\theta \\ \end{align*} \nonumber

Evaluate the inner integral \begin{align*} \int_0^{2a\cos\theta} r^2 r\,dr &=\left[\frac{r^4}{4}\right]_{r=0}^{2a\cos\theta} \\ &= 4a^4\cos^4\theta \end{align*} \nonumber

Evalutate the outer integral \begin{align*} I_o & =\iint r^2\,dA \\ &= \int_{-\pi/2}^{\pi/2} 4a^4\cos^4\theta\,d\theta = \dots = \frac{3}{2}\pi a^4 \end{align*} \nonumber

It is three times harder to spin a Frisbee about a point on a circumference than around the center.

## Change of variables

We change variables, when it simplifies the integrand or bounds, so it becomes easier to compute the double integral.

### Examples

#### One

Determine the area of an ellipse with semi-axes $$a$$ and $$b$$. $$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1 \nonumber$$

The double integral for the area $$\rm{Area} = \iint_{\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2\lt 1} dx\,dy \nonumber$$

Use substitution to make it look more like a circle \left. \begin{array}{c} \text{set }\frac{x}{a}=u \Rightarrow du = \frac{1}{a}dx \\ \text{set }\frac{y}{a}=v \Rightarrow dv = \frac{1}{b}dy \end{array} \right\} \\ \begin{align*} \Rightarrow du\,dv &= \frac{1}{ab}dx\,dy \\ \Rightarrow dx\,dy &= ab\,du\,dv \end{align*} \nonumber

Substitute it back in the double integral \begin{align*} \rm{Area} &= \iint_{u^+v^2\lt 1} ab\,du\,dv \\ &= ab\underbrace{\iint_{u^+v^2\lt 1} du\,dv}_{\text{area of unit disk}} = a\,b\,\pi \end{align*} \nonumber

#### Two

To simply integrand or bounds, we set a change of variables as \left\{ \begin{align*} u &= 3x-2y \\ v &= x+y \end{align*} \right. \nonumber

What is the relation between $$dA=dx\,dy$$ and $$dA’=du\,dv$$?

The linear transformation changes it to a parallelogram. Because of the linear change of variables, the area scaling factor doesn’t depend on the choice of rectangle. So let’s take the simplest rectangle, the unit square.

Applying the transformation to the corners

The area $$A’$$ is the determinant of the two vectors from the origin $$A’ = \left| \begin{array}{rr} 3 & 1 \\ -2 & 1 \end{array} \right| = 3+2=5 \nonumber$$

For any other rectangle, area is also multiplied by $$5$$ \begin{align*} dA’ &= 5\,dA \\ \Rightarrow du\,dv &= 5\,dx\,dy \\ \Rightarrow \iint\ldots\,dx\,dy &= \iint\ldots\,\frac{1}{5}du\,dv \end{align*} \nonumber

### Jacobian

Changing variables to $$u,v$$ means \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} u = u(x,y) &\Rightarrow \Delta u\approx \pdv{u}{x}\Delta x+\pdv{u}{y}\Delta y = u_x\Delta x+u_y\Delta y \\ v = v(x,y) &\Rightarrow \Delta v\approx \pdv{v}{x}\Delta x+\pdv{v}{y}\Delta y = v_x\Delta x + v_y\Delta y \end{align*} \right. \nonumber

In matrix form $$\left[ \begin{array}{c} \Delta u \\ \Delta v \end{array} \right] \approx \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \left[ \begin{array}{c} \Delta x \\ \Delta y \end{array} \right] \nonumber$$

A small rectangle in $$xy$$-coordinates corresponds to a small parallelogram in $$uv$$-coordinates. The sides of the parallelogram from $$(0,0)$$, are the vectors $$\left\langle\Delta x,0\right\rangle$$ and $$\left\langle 0,\Delta y\right\rangle$$ \left\{ \begin{align*} \left\langle\Delta x,0\right\rangle \rightarrow \left\langle\Delta u,\Delta v\right\rangle &\approx \left\langle u_x\Delta x, v_x\Delta x\right\rangle \\ \left\langle 0,\Delta y\right\rangle \rightarrow \left\langle\Delta u,\Delta v\right\rangle &\approx \left\langle u_y\Delta y, v_y\Delta y\right\rangle \end{align*} \right. \nonumber

The area $$\rm{Area}’$$ of the parallelogram is the determinant $$\text{Area}’ = \rm{det} \left( \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \right) \Delta x\,\Delta y \nonumber$$

When you have a general change of variables, $$du\,dv$$ versus $$dx\,dy$$ is given by the determinant of the matrix of partial derivatives. $$\rm{det}\left( \left[ \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right] \right) \nonumber$$

The definition of Jacobian just means the ratio between $$du\,dv$$ and $$dx\,dy$$. (Not a partial derivative.) Here the vertical bars stand for determinant.

$$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \shaded{ J = \pdv{(u,v)}{(x,y)} = \left| \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right| } \nonumber$$

Then, because area is always positive $$\shaded{ du\,dv = |J|\,dx\,dy = \left|\pdv{(u,v)}{(x,y)}\right|\,dx\,dy } \nonumber$$

### Examples

#### One

Switching to polar coordinates \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \end{align*} \nonumber

The Jacobian \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{(x,y)}{(r,\theta)} &= \left| \begin{array}{cc} x_r & x_\theta \\ y_r & u_\theta \end{array} \right| \\ &= \left| \begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{array} \right| \\ &= r\cos^2\theta – (-r\sin^2\theta) \\ &= r(\cos^2\theta + \sin^2\theta) = r \end{align*}

Not a constant, but a function of $$r$$, so \shaded{ \begin{align*} dx\,dy &= |r|\,dr\,d\theta \\ &= r\,dr\,d\theta \end{align*} } \nonumber

Remark: you can compute the one that easier to compute, because they are the inverse of each other. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{(u,v)}{(x,y)} \cdot \pdv{(x,y)}{(u,v)} = 1 \nonumber$$

#### Two

Compute $$\int_0^1\int_0^1 x^2y\,dx\,dy \nonumber$$

using change of variables to \left\{ \begin{align*} u &= x \\ v &= xy \end{align*} \right. \nonumber

Step 1: Find the area element using the Jacobian \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{(x,y)}{(r,\theta)} &= \left| \begin{array}{cc} u_x & u_y \\ v_x & v_y \end{array} \right| \\ &= \left| \begin{array}{cc} 1 & 0 \\ y & x \end{array} \right| = x \end{align*} With $$x$$ positive in the region \begin{align*} du\,dv &= |x|\,dx\,dy \\ &= x\,dx\,dy \end{align*} \nonumber

Step 2: Express the integrand in terms of $$u,v$$ \begin{align*} x^2y\,dx\,dy &= x^2y\,\frac{1}{x}\,du\,dv = xy\,du\,dv \\ &= u\frac{v}{u}\,du\,dv = v\,du\,dv \end{align*} \nonumber Compute (or $$dv\,du$$) $$\iint_\ldots v\,du\,dv \nonumber$$

Step 3: Find the bounds for $$u,v$$ in the new integral \begin{align*} \int_\ldots^\ldots \underbrace{\int_\ldots^\ldots v\,du}_{u \text{ changes},\\ v\text{ is constant}}\,dv \end{align*} \nonumber $$v=\rm{constant} \rightarrow xy=\rm{constant} \rightarrow y=\frac{\rm{constant}}{x}$$

What is the value of $$u$$ when we enter the region from the top, where $$y=1$$? \begin{align*} y &=1 \\ \Rightarrow y &=\frac{v}{u}=1 \\ \Rightarrow u &= v \end{align*} What is the value of $$u$$ when we exit the region, where $$x=1$$? \begin{align*} x &=1 \\ \Rightarrow u &= 1 \end{align*} The smallest value of $$(x,y)$$ is $$(0,0)$$, what corresponds to $$v=0$$. The largest value of $$(x,y)$$ is $$(1,1)$$, what corresponds to $$v=1$$.

Step 4: The double integral follows as \begin{align*} \int_0^1 \int_v^1 v\,du\,dv \end{align*} \nonumber $$\nonumber$$

How could we have found the bounds easier? Draw the picture the $$uv$$-coordinates.

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