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Mathematical Expression Editor
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
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:
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:
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
\[ R = \{(x,y):\text {$0\le x\le 1$ and $0\le y\le 1$}\}. \]
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
\[ \int _0^1 \int _0^1 x e^{xy} \d x \d y = \answer [given]{e-2} \]
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.