Iterated Integrals
We can use our understanding of partial derivatives to give some motivation to iterated integration. This is a deep and rich area. The study of integrable functions and convergence led to the modern area of mathematics known as analysis. Calculus is a small branch of modern analysis.
1. Double Integrals and Important Caveats
When we see an integral like the one given on the right, we make similar considerations to them as we did to partial derivatives. In this portion of the course, we’re largely concerned with your mechanical ability to recognize and apply the principals of integration and differentiation, but we will consider some of the underlying theory. When you see something like a double integral, a nagging voice in the back of your mind should be reminding you that integrals are defined in terms of the limit and limits rarely commute. This should immediately make you aware the double integrals will not always commute! Order often matters.
Aside from this, we simply integrate from inside to outside treating variables independent of our integration parameter as constants. Notice, that for example 1, we could have swapped the order of integration and we still arrive at the same answer!
Now, consider examples 2 and 3. They are essentially the same problem with the order of integration swapped. When we compute them, we do not arrive at the same result! What’s going on here?
If we plot this surface on wolfram alpha we obtain the figure to the right. We can see that if we move towards the origin some directions we to wards $\infty$ and in other directions we move towards $-\infty$. This integral has issues with convergence. Moreover, the integral of the absolute value of the integrand simply does not converge. This integral, therefore, is not absolutely convergent. That is $\int_a^b |f(x,y)|dxdy = \infty$.
A certain amount of cancellation allows us to obtain finite values from these integrals under the conditions of standard integration.
$\int_c^d \int_a^b f(x,y) dxdy$
Example 1: Evaluate $\int_1^2\int_1^2 \frac{1}{xy}dxdy$
$\int_1^2\left[\frac{1}{y}ln(x)\right]_1^2dy=$
$\int_1^2\frac{1}{y}ln(2)dy=$
$ln(2)\int_1^2\frac{1}{y}dy=$
$ln(2)\left[ln(y)\right]=$
$ln(2)^2$
Example 2: Evaluate $\int_0^1\int_0^1 \frac{x^2 – y^2}{(x^2 + y^2)^2}dxdy$
$\int_0^1\left[\frac{-x}{x^2 + y^2}\right]_0^1dy$
$\int_0^1\frac{-1}{1 + y^2}dy$
$\left[-arctan(y)\right]_0^1=-\frac{\pi}{4}$
Example 3: Evaluate $\int_0^1\int_0^1 \frac{x^2 – y^2}{(x^2 + y^2)^2}dydx$
$\int_0^1\left[\frac{y}{x^2 + y^2}\right]_0^1dy$
$\int_0^1\frac{1}{1 + x^2}dy$
$\left[arctan(x)\right]_0^1=\frac{\pi}{4}$
Let’s consider a few more easy examples of iterated integrals:
Example 4: Integrate $\int_0^t\int_0^s x^2 + xy + y^2 + 4 dxdy$.
$\int_0^t\int_0^s x^2 + xy + y^2 + 4 dxdy =$
$\int_0^t\left[ \frac{x^3}{3} + \frac{x^2y}{2} + y^2x + 4x\right]_0^s dy =$
$\int_0^t\frac{s^3}{3} + \frac{s^2y}{2} + y^2s + 4s dy =$
$\left[\frac{s^3y}{3} + \frac{s^2y^2}{4} + \frac{y^3s}{3} + 4sy \right]_0^t =$
$\frac{s^3t}{3} + \frac{s^2t^2}{4} + \frac{t^3s}{3} + 4st$
Example 5: Integrate $\int_0^2\int_0^1 x^2y^2 + x dxdy$.
$\int_0^2\int_0^1 x^2y^2 + x dxdy =$
$\int_0^2\left[ \frac{x^3y^2}{3} + \frac{x^2}{2}\right]_0^1 dy =$
$\int_0^2 \frac{y^2}{3} + \frac{1}{2} dy =$
$\left[\frac{y^3}{9} + \frac{y}{2} \right]_0^2 =$
$\frac{17}{9}$
2. Defining regions of integration for iterated integrals
It is very common for our region integration to be rectangular and in this case we can easily define our bounds of integration. However, it is often useful to consider nonrectangular regions of integration as well.
Example 6: Consider a function f(x, y) defined on the region depicted to the right. Write an integral over this region. Assume the order of integration does not matter.
$\int_0^2\int_0^3 f(x, y) dydx$ or $\int_0^3\int_0^2 f(x, y) dxdy$
Note that we must align the 0 to 2 bounds with dx and the 0 to 3 bounds with the dy.
Bounds could be functional though. Try the next example.
Example 7: Consider the finite region bounded by the parabola $y = x^2$ and $y = 1$. Write an integral of f(x, y) defined on the domain in the most sensible way.
This region is depicted to the right.
$\int_{-1}^1 \int_{x^2}^1 f(x, y) dydx$
Here, I want to make the point that the above way of writing the integral is usually “more correct”. For a given function f(x,y), this integral will return a value whereas reordering the bounds yields a functional result instead. There are, however, instances where you want a functional result.
Example 8: Consider the region depicted below. Construct the correct integral given $f(x , y) = x^2y$ and evaluate it.
$\int_0^2 \int_0^x x^2y dydx =$
$\int_0^2 \frac{1}{2} \left[ x^2y^2 \right]_0^xdx =$
$\int_0^2 \frac{1}{2} x^4 dx =$
$\frac{1}{10}\left[ x^5 \right]_0^2 =$
$\frac{32}{10} = \frac{16}{5}$
We can see domains need not be square. In the case of one dimensional integrals, we didn’t need to worry about the shape of our domain because everything in one dimension is a point or a line.
3. Why might you want to swap the order of integration?
Sometimes, we simply cannot swap the order integration, but we may find it advantageous to do so for a number of reasons. Sometimes integrating in a particular order yields a much easier integral. Often times, there is only a small number of orderings which yield a clear value at all from an integral! We also need to get used to conceptualizing domains of integration. Double integrals represent functions regions of or subsets of $\mathbb{R}^2 \rightarrow \mathbb{R}$.
At then end of the day, in general, an integral is just a number! Notice that, as written, example 6 will not have a closed form integral. However, it is fairly clear that this integral has a finite value on this region and we should be therefore be able to swap the bounds.
Example 9: Draw the region of integration associated with $\int_0^1\int_x^1 e^{y^2} dydx$.
Example 10: Swap the bounds of $\int_0^1\int_x^1 e^{y^2} dydx$ and rewrite them correctly using your picture from example 9. Then evaluate the integral!
$\int_0^1\int_0^y e^{y^2} dxdy$
$\int_0^1 \left[xe^{y^2}\right]_0^y dy$
$\int_0^1 ye^{y^2} dy$
$\int_0^1 ye^{y^2} dy$
$\left[\frac{e^{y^2}}{2}\right]_0^1$
$\frac{e}{2}-\frac{1}{2}$
As you can see above, we could not exactly compute our integral in example 6 as written (we could approximate it as well as we liked), but with some careful thought and rewriting, we can now obtain a closed form and a delightfully simply example!
4. Polar Integrals
Polar substitutions are extremely useful as an integration technique. This also helps us demonstrate how change of variables will work before we get to that technique. Hopefully, we have underscore that a significant portion of the problem of integration in higher dimensions comes from adequately defining the region of integration and finding a space where that area is best represented.
Recall that an iterated integral can be thought of as a reimann sum of prisms in the right number of dimensions. For instance, $\int_a^b f(x,y) dxdy$ is the sum of a collection of 3 dimensional rectangular prisms where the height is taken to be the values of $f(x, y)$ over those square bases.
We’re going to look at a slice of radial function to right. Particularly, pay attention to the green box and its labeled sides. Notice that the length a and c are approximately the same for most squares and we’re going to take a limit eventually so they will be the same provide our function is well-behaved. $a = r_{i+1} – r_i = \Delta r = c$. Now side d has length $r_i \cdot (\theta_{i+1} – \theta_i) = r_i \Delta\theta$. Similarly, the length of b is $r_{i+1} \Delta\theta$. Now, we see the area of this irregular curved quadrilateral is approximately $A \approx r_i \Delta\theta\Delta r$ and if take a limit as these become very small, we find $A = rd\theta dr. So if we call the function $f(r, \theta)$ then we could write a relevant integral as $\int_0^R \int_0^{2\pi} f(r, \theta) rd\theta dr$.
We use our standard polar substitutions to change rectangular integrals into polar integrals now. There are at least two other roads to this definition of the polar integral as well and we will explore both of them later. Let’s work a very simple example together.
Consider the region bounded by the quarter circle to the left. We will integrate $f(x) = x^2 + 2xy + y^2$ over this region. Note that the rectangular integral here would be quite difficult to accomplish symbolically. So, here we have $\int \int_D f(x) dxdy$. We can specify this more analytically if we wish to obtain:
Example 1: $\int_0^2 \int_0^{4-x^2} x^2 + 2xy + y^2 dydx$
But we can rewrite this as
$\int_0^{\pi/2} \int_0^2 ((x^2 + y^2) + 2xy) rdrd\theta =$
$\int_0^{\pi/2} \int_0^2 (r^2 + 2(r cos(\theta))(r sin(\theta))) rdrd\theta =$
$\int_0^{\pi/2} \int_0^2 r^3 + 2r^3cos(\theta)sin(\theta) drd\theta =$
$\int_0^{\pi/2} \int_0^2 r^3(1 + 2cos(\theta)sin(\theta)) drd\theta =$
$\int_0^{\pi/2} (1 + 2cos(\theta)sin(\theta)) d\theta \int_0^2 r^3 dr =$
$\int_0^{\pi/2} (1 + 2cos(\theta)sin(\theta)) d\theta \left[\frac{r^4}{4}\right]_0^2 =$
$4\int_0^{\pi/2} 1 + 2cos(\theta)sin(\theta) d\theta =$
$4\left[\theta\right]_0^{\pi/2} + 8\int_0^{\pi/2}cos(\theta)sin(\theta)) d\theta =$
Let $u = sin(\theta) \rightarrow du = cos(\theta)$
$2\pi + 8\int_0^1u du =$
$2\pi + 8\left[\frac{u^2}{2}\right]_0^1 = 2\pi + 4$
5. Multiple integrals (Triple and Beyond)
Visualizing volume and functions in higher dimensional space can be difficult. Fortunately, we can let the math lead us and the algebra and calculus work irrespective of the dimension we find ourselves in. We might ask ourselves, what really is $\int \int \int xyz dxdydz$? Here, we find that we are integrating a 3-dimensional space over another 3-dimensional space. This is essentially impossible for us to visualize, but these sorts of integrals exist everywhere in our world!
6. Cylindrical Integrals
Cylindrical coordinates essentially recognize that there is polar symmetry along some coordinate plane and extend that plane rectangularly. A
7. Spherical Integrals
Worked Problems for Iterated Integrals
1. Determine the centroid of the region defined by $x > 0, y > 1, y < 3, \text{ and } xy < 3$.
If we graph the region, we obtain the diagram given on the right. To determine this area, we evaluate the integral:
$A = \int_1^3 \int_0^{\small{\frac{3}{x}}} 1 dy dx$
$A = \int_1^3 \frac{3}{x} dx$
$A = 3ln(3)$
$\bar{x} = \frac{1}{A} \int_1^3 \int_0^{\small{\frac{3}{y}}} x dx dy$
$\bar{x} = \frac{1}{3ln(3)} \int_1^3 \frac{9}{2}\frac{1}{y^2} dy $
$\bar{x} = \frac{3}{2ln(3)} \left[ -y^-1\right]_1^3$
$\bar{x} = \frac{3}{2ln(3)} \frac{2}{3} = \frac{1}{ln3}$
$\bar{y} = \frac{1}{A} \int_1^3 \int_0^{\small{\frac{3}{y}}} y dx dy$
$\bar{y} = \frac{1}{3ln(3)} \int_1^3 \int_0^{\small{\frac{3}{y}}} y dx dy$
$\bar{y} = \frac{1}{3ln(3)} \int_1^3 3 dy$
$\bar{y} = \frac{1}{ln(3)} 2 = \frac{2}{ln(3)}$
$\left(\frac{1}{ln(3)}, \frac{2}{ln(3)}\right)$ is the centroid.
2. (Polar Integration) Evaluate $\int \int_D xy dxdy$ where $D$ is the annulus centered on the origin with inner radius 3 and outer radius 4.
$\int \int_D xy dxdy =$
$\int_3^4 \int_0^{2\pi} (r cos(\theta))(r sin(\theta)) r d\theta dr =$
$\int_0^{2\pi}\int_3^4 r^3 cos(\theta) sin(\theta) dr d\theta =$
$\int_0^{2\pi} \left[ \frac{r^4}{4} cos(\theta) sin(\theta) \right]_3^4 d\theta =$
$\frac{175}{4} \int_0^{2\pi} cos(\theta) sin(\theta) d\theta =$
Let $u = sin(\theta) \rightarrow du = cos(\theta) d\theta$
$\frac{175}{4} \int_0^{0} u du = 0$
3. (Polar Integration) Evaluate $\int \int_D e^{x^2 + y^2} dxdy$ where $D$ is the unit disk.
$\int \int_D e^{x^2 + y^2} dxdy =$
$\int_0^{2\pi} \int_0^1 e^{r^2} rdrd\theta =$
Let $u = r^2 \rightarrow du = 2rdr$
$\int_0^{2\pi} \int_0^1 \frac{1}{2} e^u dud\theta =$
$\frac{1}{2} \int_0^{2\pi} \left[ e^u \right]_0^1 d\theta =$
$\frac{1}{2} \int_0^{2\pi} (e – 1) d\theta =$
$\frac{e – 1}{2} \int_0^{2\pi} 1 d\theta =$
$\frac{e – 1}{2} \left[ \theta \right]_0^{2\pi} =$
$\frac{e – 1}{2} 2\pi =\pi (e – 1)$
4. Let a region defined by $D = \{(x,y) \ni: x^2 + y^2 \leq 4y\}$. Evaluate $\int \int_D \sqrt{x^2 + y^2}dxdy$.
We can redefine D by considering
$x^2 + y^2 = 4y \rightarrow$
$r^2 = 4rsin(\theta) \rightarrow$
$r = 4sin(\theta)$
Spend some time with the rectangular coordinates to realize this is a circular of radius 2 entirely above the x-axis centered on the point (0, 2), This will give us the bounds for polar integration.
$\int \int_D \sqrt{x^2 + y^2}dxdy$
$\int \int_D \sqrt{x^2 + y^2}rd\theta dr$
$\int \int_D (r)rd\theta dr$
$\int_0^{4sin(\theta)} \int_0^{\pi} r^2 d\theta dr$
$\int_0^{\pi} \int_0^{4sin(\theta)} r^2 drd\theta$
$\int_0^{\pi} \frac{1}{3}\left[r^3 \right]_0^{4sin(\theta)} d\theta$
$\int_0^{\pi} \frac{64}{3} sin^3(\theta) d\theta$
$\frac{64}{3} \int_0^{\pi} (1-cos^2(\theta))sin(\theta) d\theta$
Let $u = cos(\theta)$ then $du = -sin(\theta)d\theta$.
$u(0) = 1, u(\pi) = -1$.
$-\frac{64}{3} \int_1^{-1} 1 – u^2 du$
$\frac{64}{3} \int_{-1}^1 1 – u^2 du$
$\frac{64}{3} \left[u – \frac{1}{3}u^3\right]_{-1}^1$
$\frac{64}{3} \left((1 – \frac{1}{3}) – (-1 + \frac{1}{3})\right)$
$\frac{64}{3} \left(\frac{4}{3}\right)$
$\frac{256}{9}$
5. (Triple Integral) Evaluate $\int \int \int 7x dx dy dz$ bounded by the finite region enclosed by the surface $x = 2y^2 + 2z^2$ and the planes x = 0 and x = 2.
There are a variety of methods one could employ to do this. First, we’ll graph the region:
On our left, we see the region we integrating over. AGAIN, this is the region we are integrating over from different vantage points, but not the function we are integrating.
If we perform the integral as an cylindrical integral then we need to choose x as our axis of our symmetry. This implies that we will transform y and z into our radial component about this axis. r is then bounded as $0 < r < \sqrt{\frac{x}{2}}$ and x runs from 0 to 2. We perform one full rotation and that gives us our change of variables.
$\int \int \int 7x dx dy dz = $
$\int_0^{2\pi} \int_0^{\sqrt{\frac{x}{2}}} \int_0^2 7x rdx dr d\theta = $
$\int_0^{2\pi} \int_0^2 \int_0^{\sqrt{\frac{x}{2}}} 7xr dr dx d\theta = $
$\int_0^{2\pi} \int_0^2 \left[ \frac{7x}{2} r^2\right]_0^{\sqrt{\frac{x}{2}}} dx d\theta = $
$\int_0^{2\pi} \int_0^2 \frac{7}{4} x^2 dx d\theta = $
$\frac{7}{4} \int_0^{2\pi} \left[ \frac{1}{3} x^3\right]_0^2 d\theta = $
$\frac{7}{4} \int_0^{2\pi} \frac{8}{3} d\theta = $
$\frac{14}{3} \int_0^{2\pi} d\theta = $
$\frac{14}{3} 2\pi = \frac{28\pi}{3}$
5. (Triple Integral) Evaluate $\int_0^2 \int_0^{\sqrt{4-y^2}} \int_{\sqrt{3x^2 + 3y^2}}^{\sqrt{16 – x^2 -y^2}} x^2 + y^2 + z^2 dz dx dy$ . Draw the domain of integration.
How you decide to go about drawing the domain depends on many things. If you have absolute mastery of the elementary surfaces introduced earlier in three space then this part may be redundant. At this point, though, I hope you do have significant mastery of functions of one variable and we use that to our advantage here. If we examine the lower bound along the dz integration plane then we observe this function $z = \sqrt{3(x^2 + y^2}$.
If we set y = 0, then we notice this curve is a constant times the absolute value of x (recall that $|x| = \sqrt{x^2}$. This implies the graph in the xz plane is a positive absolute value curve. Similarly, in the yz plane, we get the identical curve. finally, note that in the xy plane we would fix z as some constant and then we can rewrite this as $\frac{k^2}{3} = x^2 + y^2$ implying that the xy cross sections are circles. The bottom portion of this region is bounded by a cone.
If we duplicate the same reasoning for the upper bound, $z = \sqrt{16 – x^2 -y^2}$ , we get the purple hemisphere pictured to the right. Now, we have to consider where these surfaces intersect. The intersection of the volume enclosed by these two surfaces is given to the right. It’s shaped a bit like an ice cream cone!
Next, we examine the bounds in the x direction. We see that $x = \sqrt{4-y^2}$ traces out a half circle in the righthand portion of the xy plane. If we then examine our y-direction bounding which places us between 0 and 2 in the y direction, we see this further reduces our xy footprint to the portion in the positive xy quadrant.
So, at this point, our integral is bound by the volume enclose in the figure to the right. I have made the intersections with the axial planes transparent so you can view this object more thoughtfully.
Incidentally if you want to plot this in geogebra then you can use:
If(sqrt(3x^(2) + 3y^(2)) ≤ sqrt(16 – x^(2) – y^(2)) ∧ 2 ≥ y ≥ 0 ∧ 2 ≥ x ≥ 0, sqrt(3x^(2) + 3y^(2)))
If(sqrt(3x^(2) + 3y^(2)) ≤ sqrt(16 – x^(2) – y^(2)) ∧ 2 ≥ y ≥ 0 ∧ 2 ≥ x ≥ 0, sqrt(16 – x^(2) – y^(2)))
Now, we get to setting up our integral. We can the radius of our hemisphere is 4 so $0 \leq \rho \leq 4$ and we are in the positive octant so $0 \leq \theta \leq \frac{\pi}{2}$. Now, we need to figure out $\phi$. It is bounded by the cone. There’s a variety of ways to find this, but I’ll focus on the most straight forward: algebraically:
We again examine the equation of the cone:
$z = \sqrt{3(x^2 + y^2)}$
and now we apply the spherical substitutions to obtain
$\rho cos(\phi) = \sqrt{ 3(\rho^2 cos^2(\theta)sin^2(\phi) + \rho^2 sin^2(\theta)sin^2(\phi))} $
$\rho cos(\phi) = \sqrt{ 3\rho^2sin^2(\phi)} $
$\rho cos(\phi) = \rho sin(\phi) \sqrt{ 3} $
$cos(\phi) = sin(\phi) \sqrt{ 3} $
$cot(\phi) = \sqrt{ 3} $
$\phi = \frac{\pi}{6}
So we have $0 \leq \phi \leq \frac{\pi}{6}$.
$\int_0^4 \int_0^{\frac{\pi}{6}} \int_0^{\frac{\pi}{2}} x^2 + y^2 + z^2 \rho^2 sin(\phi) d\theta d\phi d\rho =$
$\int_0^4 \int_0^{\frac{\pi}{6}} \int_0^{\frac{\pi}{2}} \rho^2 \rho^2 sin(\phi) d\theta d\phi d\rho =$
$\int_0^4 \int_0^{\frac{\pi}{6}} \int_0^{\frac{\pi}{2}} \rho^4 sin(\phi) d\theta d\phi d\rho =$
$\int_0^4 \rho^4 d\rho \int_0^{\frac{\pi}{6}} sin(\phi) d\phi \int_0^{\frac{\pi}{2}} d\theta =$
$\left[\frac{rho^5}{5}\right]_0^4 \left[-cos(\phi)\right]_0^{\frac{\pi}{6}} \left[\theta\right]_0^{\frac{\pi}{2}} =$
$\frac{2048}{5} (\frac{1}{2}) (\frac{\pi}{2}) = \frac{512\pi}{5}$