Iterated Integrals and Fubini's Theorem

Iterated Integrals

We are about to look at a method of evaluating double integrals over rectangles and more general domains without using the definition of a double integral, but before we do so, we will first need to learn what an iterated integral.

Suppose that $z=f(x,y)$ is a two variable real-valued function, and suppose that $f$ is integrable for $a\leq x\leq b$ and $c\leq y\leq d$, that is $f$ is integrable over the rectangle:

${\begin{aligned}\quad R=\{(x,y)\in \mathbb {R} ^{2}:a\leq x\leq b,c\leq y\leq d\}=[a,b]\times [c,d]\end{aligned}}$

Just like partial differentiation with respect to a specific variable, we can also partial integrate with respect to a specific variable. For example, $\int _{c}^{d}f(x,y)\,dy$ means that we integrate $f$ with respect to $y$ while holding the variable $x$ as fixed. When we evaluate this integral, we will obtain a function in terms of $x$ only, and hence, we could then integrate the result from $a$ to $b$ with respect to $x$ as:

${\begin{aligned}\quad \int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx\end{aligned}}$

The result above is what we call an iterated integral which we will define below.

Definition: If $z=f(x,y)$ is a two variable real valued function, and if $f$ is integrable over the rectangle $R=[a,b]\times [c,d]$, then the Iterated Integral of $f$ over $R$ is

$\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx$.

We will see later that iterated integrals need not be over rectangles but instead can be done over more general domains.

One important property about iterated integrals is that we can partially integrate $f(x,y)$ with respect to either variable $x$ or $y$ first, and then continue onward with integrating with respect to the second variable, that is:

${\begin{aligned}\quad \int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx=\int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy\end{aligned}}$

However, it is important to note that sometimes partial integrating with respect to a certain variable first will be a much easier process. Let's now look at some examples of evaluating iterated integrals.

Example 1

Evaluate the following iterated integral $\int _{2}^{4}\int _{1}^{3}x^{3}+xy^{2}\,dy\,dx$. Over what rectangle is $f(x,y)=x^{3}+xy^{2}$ being integrated?

When evaluated iterated integrals over rectangles, we always want to work from the inside out. Let's first evaluate the inside integral with respect to $y$ while holding $x$ as fixed.

${\begin{aligned}\quad \int _{1}^{3}x^{3}+xy^{2}\,dy=\left[x^{3}y+{\frac {xy^{3}}{3}}\right]_{1}^{3}=\left(3x^{3}+9x\right)-\left(x^{3}+{\frac {x}{3}}\right)=2x^{3}+{\frac {26x}{3}}\end{aligned}}$

Therefore we have that:

${\begin{aligned}\quad \int _{2}^{4}\int _{1}^{3}x^{3}+xy^{2}\,dy\,dx=\int _{2}^{4}2x^{3}+{\frac {26x}{3}}\,dx\end{aligned}}$

Evaluating this definite integral and we get that:

${\begin{aligned}\quad \int _{2}^{4}2x^{3}+{\frac {26x}{3}}\,dx=\left[{\frac {x^{4}}{2}}+{\frac {26x^{2}}{6}}\right]_{2}^{4}=\left(128+{\frac {416}{6}}\right)-\left(8+{\frac {104}{26}}\right)=120+52=172\end{aligned}}$

Therefore we have that:

${\begin{aligned}\quad \int _{2}^{4}\int _{1}^{3}x^{3}+xy^{2}\,dy\,dx=172\end{aligned}}$

In this particular example, we are integrating over the rectangle $R=[2,4]\times [1,3]$.

Example 2

Evaluate the following iterated integral $\int _{0}^{\pi }\int _{0}^{\pi }\sin x\cos y\,dx\,dy$. Over what rectangle is $f(x,y)=\sin x\cos y$ being integrated?

We will first start by evaluating the inner integral while holding $y$ as fixed:

${\begin{aligned}\quad \int _{\frac {\pi }{2}}^{\pi }\sin x\cos y\,dx=\left[-\cos x\cos y\right]_{0}^{\pi }=\left(\cos y\right)-\left(-\cos y\right)=2\cos y\end{aligned}}$

Therefore we have that:

${\begin{aligned}\quad \int _{0}^{\pi }\int _{0}^{\pi }\sin x\cos y\,dx\,dy=\int _{0}^{\pi }2\cos y\,dy\end{aligned}}$

Evaluating this definite integral and we get that:

${\begin{aligned}\quad \int _{0}^{\pi }2\cos y\,dy=\left[2\sin y\right]_{0}^{\pi }=0\end{aligned}}$

In this particular example, we are integrating over the square $R=[0,\pi ]\times [0,\pi ]$.

Fubini's Theorem and Evaluating Double Integrals over Rectangles

We have just looked at Iterated Integrals over rectangles. You might now wonder how iterated integrals relate to double integrals that we looked are earlier. Fubini's Theorem gives us a relationship between double integrals and these iterated integrals.

Theorem 1 (Fubini's Theorem): Let $z=f(x,y)$ be a two variable real-valued function. If $f$ is continuous on the rectangle $R=[a,b]\times [c,d]$ then the double integral over $R$ can be computed as an iterated integrals and $\iint _{R}f(x,y)\,dA=\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx=\int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy$.

Fubini's Theorem is critically important as it gives us a method to evaluate double integrals over rectangles without having to use the definition of a double integral directly.

Now the following corollary will give us another method for evaluating double integrals over a rectangle $R=[a,b]\times [c,d]$ provided that $f$ can be written as a product of a function in terms of $x$ and a function in terms of $y$.

Corollary 1: Let $z=f(x,y)$ be a two variable real-valued function. If $f(x,y)=g(x)h(y)$ and $f$ and $R=[a,b]\times [c,d]$ then $\iint _{R}f(x,y)\,dA=\iint _{R}g(x)h(y)\,dA=\left[\int _{a}^{b}g(x)\,dx\right]\left[\int _{c}^{d}h(y)\,dy\right]$.

Example 1

Evaluate $\iint _{R}xy+y^{2}\,dA$ where $R=[0,1]\times [1,2]$.

By Fubini's Theorem we have that:

${\begin{aligned}\quad \iint _{R}xy+y^{2}\,dA=\int _{0}^{1}\int _{1}^{2}xy+y^{2}\,dy\,dx\end{aligned}}$

Now let's evaluate the inner integral $\int _{1}^{2}xy+y^{2}\,dy$ first while holding $x$ as fixed:

${\begin{aligned}\quad \int _{1}^{2}xy+y^{2}\,dy=\left[{\frac {xy^{2}}{2}}+{\frac {y^{3}}{3}}\right]_{1}^{2}=\left(2x+{\frac {8}{3}}\right)-\left({\frac {x}{2}}+{\frac {1}{3}}\right)={\frac {3x}{2}}+{\frac {7}{3}}\end{aligned}}$

And so we have that:

${\begin{aligned}\quad \int _{0}^{1}\int _{1}^{2}xy+y^{2}\,dy\,dx=\int _{0}^{1}{\frac {3x}{2}}+{\frac {7}{3}}\,dx=\left[{\frac {3x^{2}}{4}}+{\frac {7x}{3}}\right]_{0}^{1}={\frac {3}{4}}+{\frac {7}{3}}={\frac {37}{12}}\end{aligned}}$

Example 2

Evaluate $\iint _{R}e^{x}\cos y\,dA$ where $R=[0,1]\times \left[{\frac {\pi }{2}},\pi \right]$.

Note that $f(x,y)=e^{x}\cos y$ can be written as the product of a function of $x$ and a function of $y$ if we let $g(x)=e^{x}$ and $h(y)=\cos y$ (then $f(x,y)=g(x)h(y)$). Therefore applying Corollary 1 and we get that:

${\begin{aligned}\quad \iint _{R}e^{x}\cos y\,dA=\left[\int _{0}^{1}e^{x}\,dx\right]\left[\int _{\frac {\pi }{2}}^{\pi }\cos y\,dy\right]=[e-1][-1]=1-e\end{aligned}}$

Fubini's Theorem for Evaluating Triple Integrals over Boxes

We will now look at an analogous Theorem for evaluating triple integrals over a rectangular box $B$ with iterated integrals.

Theorem 1: Let $w=f(x,y,z)$ be a three variable real-valued function such that $f$ is continuous on the rectangular box $B=[a,b]\times [c,d]\times [r,s]$. Then the triple integral of $f$ over $B$ can be evaluated as iterated integrals, that is, $\iiint _{B}f(x,y,z)\,dV=\int _{r}^{s}\int _{c}^{d}\int _{a}^{b}f(x,y,z)\,dx\,dy\,dz$.

There are six total ways to evaluate a triple integral over a box using iterated integrals. For example, we could get that $\iiint _{B}f(x,y,z)\,dV=\int _{a}^{b}\int _{c}^{d}\int _{r}^{s}f(x,y,z)\,dz\,dy\,dx$, or $\iiint _{B}f(x,y,z)\,dV=\int _{c}^{d}\int _{a}^{b}\int _{r}^{s}f(x,y,z)\,dz\,dx\,dy$. Each of these six possible orders will give rise to the same value.

Let's now look at an example of evaluating a triple integral over a box.

Example 1

Evaluate the triple integral $\iiint _{B}2x-e^{y}+3z^{2}\,dV$ over the box $B=[0,2]\times [0,1]\times [1,3]$.

By Fubini's Theorem we can rewrite this triple integral as iterated integrals:

${\begin{aligned}\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\int _{1}^{3}\int _{0}^{1}\int _{0}^{2}2x-e^{y}+3z^{2}\,dx\,dy\,dz\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\int _{1}^{3}\int _{0}^{1}\left[x^{2}-xe^{y}+3xz^{2}\right]_{0}^{2}\,dy\,dz\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\int _{1}^{3}\int _{0}^{1}4-2e^{y}+6z^{2}\,dy\,dz\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\int _{1}^{3}\left[4y-2e^{y}+6yz^{2}\right]_{0}^{1}\,dz\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\int _{1}^{3}6-2e+6z^{2}\,dz\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=\left[6z-2ez+2z^{3}\right]_{1}^{3}\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=(18-6e+54)-(6-2e+2)\\\quad \iiint _{B}2x-e^{y}+3z^{2}\,dV=64-4e\end{aligned}}$

Resources

• Iterated Integrals, Paul's Online Notes

Licensing

Content obtained and/or adapted from:

• Iterated integrals, mathonline.wikidot.com under a CC BY-SA license
• Fubini's Theorem and Evaluating Double Integrals over Rectangles, mathonline.wikidot.com under a CC BY-SA license
• Fubini's Theorem for Evaluating Triple Integrals over Boxes, mathonline.wikidot.com under a CC BY-SA license