We study integrals over basic regions.
As we journey through calculus again, we are now ready for integrals. Instead of
integrating over an interval like \([a,b]\) we now integrate over regions like this
\[ R = \{(x,y):\text {$x$ and $y$ satisfy some property}\} \]
in particular,
in this section we will only consider rectangles and boxes, and this is what we mean
by “trivial” regions. Here a rectangle is defined as:
\[ R = \{(x,y):\text {$a\le x\le b$ and $c\le y\le d$}\} \]
A box is defined as:
\[ B = \{(x,y,z):\text {$a\le x\le b$, $c\le y\le d$, and $p\le z\le q$}\} \]
Let’s get to
work!
1 Double integrals
Suppose you have a function \(F:\R ^2\to \R \). A graph of this function is a surface in \(\R ^3\). For
example:
We are interested in the “signed volume” or “net volume” between this surface and
the \((x,y)\)-plane. This means that the space above the \((x,y)\) plane under \(F\) will have a positive
volume. Space above \(F\) and under the \((x,y)\)-plane will have a “negative” volume. This is
similar to the notion of “signed area” used before. If we want to compute the signed
volume of a surface defined by \(F\) over a rectangular region, say the rectangle
defined by
\[ R = \{(x,y):\text {$a\le x\le b$ and $c\le y\le d$}\} \]
we break \(R\) into \(n\) slices parallel to the \(x\)-axis, and \(m\) slices parallel to
the \(y\)-axis. This allows us to consider boxes of dimension
\[ \underbrace {F(x_i^*,y_j^*)}_{\text {height}}\times \underbrace {\left (\frac {b-a}{m}\right )}_{\Delta x}\times \underbrace {\left (\frac {d-c}{n}\right )}_{\Delta y} \]
where \((x_i^*,y_j^*)\) is a point
in \((i,j)\)-rectangle:
Computing the volume of each of these boxes approximates and summing them
together
\[ \sum _{j=1}^n\sum _{i=1}^m F(x^*_i,y^*_j)\Delta x\Delta y \]
approximates the signed volume enclosed by the surface:
Letting the number of rectangles in the \(x\)-direction and \(y\)-direction go to infinity, we
will have that \(\Delta x \cdot \Delta y\) goes to zero, and we will find the exact volume enclosed by our
surface when bounded by the region \(R\). This leads to our definition of a double
integral:
Given a function
\(F:\R ^2\to \R \), a
double integral \[ \iint _R F(x,y) \d A \]
of a function
\(F\) over a rectangular region
\(R\), is
given by:
\[ \iint _R F(x,y) \d A = \lim _{\substack {m\to \infty \\ n\to \infty }}\sum _{j=1}^n\sum _{i=1}^m F(x^*_i,y^*_j)\Delta x\Delta y \]
Let the value of a function
\(F:\R ^2\to \R \) be given below:
Let
\[ R = \{(x,y): \text {$0\le x\le 4$ and $0\le y\le 6$}\} \]
\begin{align*} \Delta x &= \answer {1}\\ \Delta y &= \answer {2} \end{align*}
Now, simply add up the \(z\)-values found in the rectangles that are within our region
and multiply by \(\answer {2}\). Compute \(\iint _R F(x,y)\d A\).
\[ \iint _R F(x,y) \d A = \answer {28} \]
How do we compute a double integral with calculus? We use an iterated integral. At
this point we will introduce something called Fubini’s Theorem.
1.1 Fubini’s Theorem
Fubini’s Theorem gives us a recipe for computing double integrals. In this class, we
are going to have many different versions of Fubini’s Theorem. The common factor
between all of these theorems is that with each, the “punch-line” will be:
\[ \underbrace {\iint _R F(x,y) \d A}_{\text {a double integral}} = \underbrace {\int _?^? \int _?^? F(x,y) \d A}_{\text {an iterated integral}} \]
where an
iterated integral is nothing more than two applications of our familiar friend/foe: the
single integral.
Fubini’s Theorem Let
\(F\) be continuous on the region
\[ R = \{(x,y):\text {$a\le x\le b$ and $c\le y\le d$}\}. \]
Then:
\begin{align*} \iint _R F(x,y) \d A &= \int _a^b\int _c^d F(x,y)\d y\d x\\ &=\int _c^d\int _a^b F(x,y)\d x\d y. \end{align*}
Now let’s work some examples:
Let
\(F(x,y) = xy+e^y\). Find the signed volume under
\(F\) on the region
\[ R = \{(x,y):\text {$3\le x\le 4$ and $1\le y\le 2$}\}. \]
![[Picture]](digInIntegralsOverTrivialRegions.online-727b1c684128b2d427e68ec00a5b5871.svg)
We will compute
this integral two different ways. First apply Fubini’s Theorem:
\[ \iint _R \big (xy+e^y\big ) \d A = \int _3^{\answer [given]{4}}\int _{\answer [given]{1}}^{\answer [given]{2}}\big (xy+e^y\big ) \d \answer [given]{y} \d \answer [given]{x} \]
Now integrate with
respect to
\(y\), treating
\(x\) as a constant:
\begin{align*} = \int _{\answer [given]{3}}^{\answer [given]{4}} \eval {\answer [given]{\frac {1}{2}xy^2+e^y}}_{\answer [given]{1}}^{\answer [given]{2}} \d \answer [given]{x} \\ &= \int _{\answer [given]{3}}^{\answer [given]{4}}\left (\answer [given]{\frac {3}{2}x + e^2-e}\right )\d \answer [given]{x} \end{align*}
Now integrate with respect to \(x\), treating \(y\) as a constant:
\begin{align*} &= \eval {\answer [given]{\frac {3}{4}x^2 + (e^2-e)x}}_{\answer [given]{3}}^{\answer [given]{4}} \\ &= \answer [given]{\frac {21}{4}+ e^2-e}. \end{align*}
Now let’s compute this integral using a different order of integration. Write with
me,
\begin{align*} \iint _R\big (xy+e^y\big ) \d A &= \int _1^{\answer [given]{2}}\int _{\answer [given]{3}}^{\answer [given]{4}}\big (xy+e^y\big )\d \answer [given]{x} \d \answer [given]{y} \\ &= \int _{\answer [given]{1}}^{\answer [given]{2}}\eval {\answer [given]{\frac {1}{2}x^2y+xe^y}}_{\answer [given]{3}}^{\answer [given]{4}}\d \answer [given]{y}\\ &= \int _{\answer [given]{1}}^{\answer [given]{2}}\left (\answer [given]{\frac {7}{2}y+e^y}\right )\d \answer [given]{y}\\ &= \eval {\answer [given]{\frac {7}{4}y^2+e^y}}_{\answer [given]{1}}^{\answer [given]{2}}\\ &=\answer [given]{\frac {21}{4}+e^2-e}. \end{align*}
Note our answers are the same regardless of the order of integration.
In our next example, we will see that it is sometimes easier to apply Fubini’s
Theorem and integrate with respect to one variable instead of the other.
Let
\(F(x,y) = xe^{xy}\). Find the signed volume under
\(F\) on the region
\[ R = \{(x,y):\text {$0\le x\le 1$ and $0\le y\le 1$}\}. \]
![[Picture]](digInIntegralsOverTrivialRegions.online-d16572ffb910fad3d98ef9287f5d8027.svg)
Let’s first
compute:
\[ \int _0^1 \int _0^1 x e^{xy} \d x \d y \]
As we will see, this integral will require three tricks. To integrate with
respect to
\(x\), you can use integration by parts:
\begin{align*} \int _0^1 \int _0^1 x e^{xy} \d x \d y &= \int _0^1 \eval {\answer [given]{\frac {xe^{xy}}{y}-\frac {e^{xy}}{y^2}}}_0^1\d y\\ &= \int _0^1 \left (\frac {e^{y}}{y}-\frac {e^y}{y^2} + \frac {1}{y^2}\right )\d y \end{align*}
Now we notice that this is an improper integral! We must therefore compute
\[ \lim _{b\to \answer [given]{0}}\int _{\answer [given]{b}}^{\answer [given]{1}} \left (\answer [given]{\frac {e^{y}}{y}-\frac {e^y}{y^2} + \frac {1}{y^2}}\right )\d y \]
Now we
must integrate with respect to
\(y\). This is also tricky. Our second trick is to rewrite our
current integrand as
\[ \frac {y e^y - \left (\answer [given]{e^y -1}\right )}{y^2} \]
and now “see” that this results from the quotient rule being
applied to
\[ \frac {e^y-\answer [given]{1}}{y}. \]
So now we must compute:
\begin{align*} \lim _{b\to \answer [given]{0}}\eval {\frac {e^y-1}{y}}_{\answer [given]{b}}^{\answer [given]{1}} &= \lim _{\answer [given]{b}\to \answer [given]{0}}\left (\answer [given]{\frac {e^1-1}{1}-\frac {e^b-1}{b}}\right )\\ &= \frac {e^1-1}{1}-\lim _{b\to 0}\answer [given]{\frac {e^b-1}{b}}. \end{align*}
Now, for our third trick, we’ll use L’Hôpital’s rule. Note that the numerator and
denominator both go to zero and are differentiable, so this is OK.
\begin{align*} \lim _{b\to 0}\frac {e^b-1}{b} &= \lim _{b\to 0}\answer [given]{\frac {e^b}{1}}\\ &= \answer [given]{1}. \end{align*}
So putting this all together, we have
\[ \int _0^1 \int _0^1 x e^{xy} \d x \d y = \answer [given]{e-2} \]
Whew. This was hard!
Now use Fubini’s Theorem and integrate with respect to \(y\) first! Fubini’s theorem says:
\[ \int _0^1 \int _0^1 x e^{xy} \d x \d y = \int _0^1 \int _0^1 x e^{xy} \d y \d x \]
provided you switch the order of integration. Note, in this case the limits of
integration for both
\(x\) and
\(y\) are the same, but we
did switch them. Let’s get on with it!
Write with me:
\begin{align*} \int _0^1 \int _0^1 x e^{xy} \d y \d x &= \int _0^1 \eval {\answer [given]{e^{xy}}}_0^1\d x\\ &= \int _0^1 \left (\answer [given]{e^x - 1}\right ) \d x \\ &= \eval {\answer [given]{e^x - x}}_0^1\\ &= \answer [given]{e-2} \end{align*}
Man alive! That was easy! Note, we get the same final answer regardless of the order
of integration. Thank-you Fubini!
2 Triple integrals
Using a similar technique to how we made boxes to define double integrals, we can
make four-dimensional boxes to define a triple integral that computes the signed
hypervolume bounded by a hypersurface and a three-dimensional region.
\[ \iiint _B F(x,y,z) \d V = \lim _{\substack {\l \to \infty \\ m\to \infty \\ n\to \infty }} \sum _{k=1}^n \sum _{j=1}^m \sum _{i=1}^\l F(x^*_i,y^*_j,z_k^*)\Delta x\Delta y\Delta z \]
How do
we compute a triple integral with calculus? We use our second version of Fubini’s
Theorem. This time, it will say something like:
\[ \underbrace {\iiint _B F(x,y,z) \d V}_{\text {a triple integral}} = \underbrace {\int _?^? \int _?^? \int _?^? F(x,y,z) \d V}_{\text {an iterated integral}} \]
Fubini’s Theorem Let
\(F\) be continuous on the region
\[ B = \{(x,y,z):\text {$a\le x\le b$, $c\le y\le d$, $p\le z\le q$}\}. \]
Then:
\[ \iiint _B F(x,y,z) \d V = \int _a^b\int _c^d\int _p^q F(x,y,z)\d z \d y\d x. \]
Here, all
six combinations
of orders of integration will work, and be equal, provided that the bounds are
changed appropriately.
Let
\(F\) be a continuous function on the region
\(B\).
\begin{align*} \int _a^b\int _c^d\int _p^q &F(x,y,z)\d z \d y\d x \\ &= \int _{\answer {p}}^{\answer {q}}\int _a^{\answer {b}} \int _{\answer {c}}^d F(x,y,z)\d \answer {y} \d \answer {x} \d \answer {z} \end{align*}
\[ \int _a^b f(x) \d x = - \int _b^a f(x) \d x \]
\begin{align*} \int _a^b\int _c^d\int _p^q &F(x,y,z)\d z \d y\d x \\ &= \int _{\answer {c}}^{d} \int _{\answer {b}}^{a}\int _{\answer {q}}^{\answer {p}} F(x,y,z)\d \answer {z} \d \answer {x} \d \answer {y} \end{align*}
Let’s do an example.
Let
\(F(x,y,z) = \cos (x-y)\sin (z)\). Find the signed hypervolume “under”
\(F\) on the region
\[ B = \{(x,y,z):\text {$0\le x\le \pi $, $0\le y\le \pi $, and $0\le z\le \pi $}\}. \]
First apply Fubini’s
Theorem to convert the triple integral into an iterated integral:
\[ \iiint _B F(x,y,z) \d V =\int _{\answer [given]{0}}^{\answer [given]{\pi }}\int _{\answer [given]{0}}^{\answer [given]{\pi }}\int _{\answer [given]{0}}^{\answer [given]{\pi }} \answer [given]{\cos (x-y)\sin (z)} \d x \d y \d z \]
Now integrate with
respect to
\(x\):
\begin{align*} &= \int _{\answer [given]{0}}^{\answer [given]{\pi }}\int _{\answer [given]{0}}^{\answer [given]{\pi }}\eval {\answer [given]{\sin (x-y)\sin (z)}}_{\answer [given]{0}}^{\answer [given]{\pi }}\d y \d z \\ &= \int _{\answer [given]{0}}^{\answer [given]{\pi }}\int _{\answer [given]{0}}^{\answer [given]{\pi }}\left (\answer [given]{\sin (\pi -y)\sin (z) + \sin (y)\sin (z)}\right )\d y \d z \end{align*}
Now integrate with respect to \(y\):
\begin{align*} &= \int _{\answer [given]{0}}^{\answer [given]{\pi }}\eval {\answer [given]{\cos (\pi -y)\sin (z) -\cos (y)\sin (z)}}_{\answer [given]{0}}^{\answer [given]{\pi }} \d z\\ &= \int _{\answer [given]{0}}^{\answer [given]{\pi }}\answer [given]{4\sin (z)} \d z \end{align*}
Finally integrate with respect to \(z\):
\begin{align*} &= \eval {\answer [given]{-4\cos (z)}}_{\answer [given]{0}}^{\answer [given]{\pi }} \\ &= \answer [given]{8} \end{align*}
We’ve just begun our journey with multiple integrals. Next, we’ll think about more
complex regions! For some interesting extra reading check out:
![[Picture]](digInIntegralsOverTrivialRegions.online-d2b7847cb919b0307a931a72833f045d.svg)
![[Picture]](digInIntegralsOverTrivialRegions.online-ec51fb62eee32e3528a33bdba095d7fe.svg)
![[Picture]](digInIntegralsOverTrivialRegions.online-fdb1c18279999687980a285449da20c3.svg)
![[Picture]](digInIntegralsOverTrivialRegions.online-98ff138489c5f46e61412607675c02ab.svg)