]> IB HL2 - Chapter 06

Applications of Integration

6.1: Area Between Two Curves

Simple! Just subtract the areas under each of the curves…or, equivalently, find the area under the curve $(f - g) (x)$ .

Be careful! Area is positive. Depending on which function is above the other, you may need to introduce an absolute value (or change the order of subtraction) to make the integral equal the area.

Example

[1.] Find the area of the region in the first quadrant bounded by the functions $f (x) = x^{2}$ and $g (x) = x$ .

Perhaps we should look at the picture first…

An Image

So, we need to subtract $g (x) - f (x)$ in order to find the area. The intersections are at zero and one, so they will become the limits of integration.

$A = \int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} x^{2} - \frac{1}{3} x^{3}]_{0}^{1} = (\frac{1}{2} {(1)}^{2} - \frac{1}{3} {(1)}^{3}) - (0) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ .

6.2: Volume (Disk Method)

The Idea

There are many shapes that are created by rotating a simple shape in space. It turns out that the volume of such objects can be found using calculus.

Here's the idea: take a small interval along the axis of rotation, and use that to create a rectangle (much like we did when beginning integrals). Now, spin that rectangle around the axis to create a very small cylinder. Knowing the volume of that cylinder, and being able to add up a lot of those cylinders means that we can find the volume of a solid that is generated by rotating a function around an axis.

The Disk Method

Let's consider a function (it doesn't matter which—call it $f (x)$ ) from $x = 0$ to $x = 4$ . Here's the graph:

An Image

When this is rotated around the x-axis, the resulting shape is…

An Image

How do we get the volume of this thing? By slicing it up into manageable parts, and taking the limit.

OK, let's consider a really tiny sliver along the x-axis and make a little rectangle…like this:

An Image

If we take this little rectangle, and spin it around the x-axis, we get a little cylinder (really short cylinders are called disks):

An Image

The volume of a cylinder/disk is simple to calculate: $V = π r^{2} h$ . The radius is just the function value ( $f (x)$ ), and the height is tiny (let's call it $Δ x$ ).

An Image

So—the volume of this small cylinder is $π {(f (x))}^{2} Δ x$ . Now, I want to add up the volumes of a bunch of these disks (specifically, from $x = 0$ to $x = 4$ ) and take the limit as $Δ x$ approaches zero. That should give the volume of this object! $V = lim_{Δ x \to 0} \sum_{x = 0}^{4} π {(f (x))}^{2} \cdot Δ x$

Now wait just a minute…this looks familiar. Haven't we done this before?

Yes! That looks like an integral!

And it is.

The volume of the object is $V = \int_{0}^{4} π {(f (x))}^{2} d x$ . Of course, you don't always have to start at zero and stop at four…you can use any reasonable limits for the integration.

This example assumes a rotation around the x-axis. If you want to rotate around the y-axis, then you probably want to rewrite the function so that x is a function of y. We'll see an example of that a bit later.

The Washer Method

Sometimes, the area that you want to rotate doesn't come out solid—it has a hole of some type. This arises when the area that is rotated is the area between two functions, rather than the area beneath just one function.

Consider these functions:

An Image

When that very unusual area is rotated, it looks like this:

An Image

(it is difficult to see, perhaps, that this thing is empty inside…try!)

We can still consider a very small section along the axis to create a rectangle. However, when this is rotated around the axis, you get what most would recognize as a ring (often called a washer)!

An Image

The volume of this ring is not difficult to find…

An Image

It's just the volume of the disk with the larger radius, minus the volume of the disk with the smaller radius. $V = π {(f (x))}^{2} Δ x - π {(g (x))}^{2} Δ x = π [{(f (x))}^{2} - {(g (x))}^{2}] Δ x$

Again, we want to sum these up and take the limit as $Δ x$ approaches zero…and again, this results in an integral!

$V = \int_{a}^{b} π [{(f (x))}^{2} - {(g (x))}^{2}] d x$

Cross Sections

In general, if we know the cross sectional area of a solid at any point along an axis (typically the x-axis), then we can find the volume of the object by integrating the area function along the axis.

Examples

[2.] Find the volume of the solid formed when $y = x^{2}$ ( $x \in [0, 5]$ ) is rotated about the x-axis.

Here's what that looks like:

An Image

And here's a typical disk formed by that rotation:

An Image

The volume is $\int_{0}^{5} π {(x^{2})}^{2} d x = \int_{0}^{5} π x^{4} d x = \frac{π}{5} x^{5}]_{0}^{5} = \frac{π}{5} (3125 - 0) = 625 π$ .

[3.] Find the volume of the solid when the region bounded by $y = x^{2}$ , the y-axis and $y = 4$ is rotated about the y-axis.

First, a picture of what that looks like:

An Image

So a typical disk looks like this:

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Note that since the disk is perpendicular to the y-axis, the thickness must be in terms of y ( $Δ y$ ). That means that we are going to have to integrate with respect to y…which means that the functions needs to define x in terms of y! Fortunately, that's not too hard…if $y = x^{2}$ , then $\pm \sqrt{y} = x$ . We need not consider both of these (we can ignore the ±) because that just mirrors the function on the other side of the y-axis (making the complete parabola shape). We need only consider/rotate one part of that: $\sqrt{y} = x$ . The typical disk has radius $\sqrt{y}$ and thickness $Δ y$ , so the volume of a typical disk is $π {(\sqrt{y})}^{2} Δ y$ . The volume of the entire solid (from $y = 0$ to $y = 4$ ) is $\int_{0}^{4} π {(\sqrt{y})}^{2} d y = \int_{0}^{4} π y d y = \frac{π}{2} y^{2}]_{0}^{4} = \frac{π}{2} (16 - 0) = 8 π$ .

(note that ${(\sqrt{y})}^{2}$ since we are only considering positive values of y)

[4.] Find the volume of the solid formed by rotating the region bounded by $y = sin x$ and $y = cos x$ (where x > 0)around the x-axis.

First of all, what does that region look like?

An Image

Note that $y = cos x$ is the function in red, and that the limits on x are $x \in [0, \frac{π}{4}]$ .

Now, the volume of revolution:

An Image

(looks like a dog's water bowl to me). Finally, a typical ring:

An Image

So the volume is $\int_{0}^{\frac{π}{4}} ({cos}^{2} x - {sin}^{2} x) d x$ . But, in order to actually work that out, you're going to have to use an identity! $cos (2 x) = {cos}^{2} x - {sin}^{2} x$ in particular. So the volume is $\int_{0}^{\frac{π}{4}} π cos (2 x) d x = \frac{π}{2} sin (2 x)]_{0}^{\frac{π}{4}} = \frac{π}{2} (sin \frac{π}{2} - sin 0) = \frac{π}{2}$ .

6.3: Volume (Shell Method)

The Idea

Sometimes, changing the variable of integration just isn't possible (or desirable). There is another method that will allow you to find the volume of revolution anyway. Note that this topic is NOT part of the IB curriculum!

If you can't change the variable of integration, then the rectangle that you create will not make a disk when rotated…it will make a thin cylindrical shell (can). Finding the volume of a can is easy; all that must then be done is to sum up a bunch of those volumes, and take the limit as the thickness of the cans approaches zero. Easy, right!

The Shell Method

Let's consider a function ( $f (x)$ ) to be rotated around the y-axis.

An Image

When this is rotated around the y-axis, you get a strange bundt-cake shape:

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A tiny sliver of x ( $Δ x$ ), when rotated, gives a thin can/shell:

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Finding the volume of this can/shell is easy!

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Well, it's easy once you see how it's done. If you cut that shell down the long way, and unroll it, you get a slab. The thickness of the slab is $Δ x$ , and its width is $f (x)$ . The length of the slab comes from the circumference of the can/shell…which is $C = 2 π r = 2 π x$ . So, the volume of the slab (can/shell) is $V = (2 π x) (f (x)) Δ x$ . Now, do that for a bunch of values of x, and take the limit as $Δ x$ approaches zero, and you get $V = \int_{a}^{b} 2 π x \cdot f (x) d x$ .

This example used a rotation around the y-axis…for a rotation around the x-axis, you'll need to write the equation of the generating curve as a function of y.

Example

[5.] Find the volume of the solid generated by rotating the region bounded by $y = sin (x^{2})$ and the x-axis about the y-axis.

First, a picture of the region:

An Image

Note that the limits on x are $x \in [0, \sqrt{π}]$ . Now, the solid:

An Image

Now, a typical shell:

An Image

So the volume is $\int_{0}^{\sqrt{π}} 2 π x \cdot sin (x^{2}) d x$ . You'll need to do a substitution to find this integral… $u = x^{2} ⇒ d u = 2 x d x ⇒ \frac{d u}{2} = x d x$ . So $\int_{0}^{\sqrt{π}} 2 π x \cdot sin (x^{2}) d x = π \int_{0}^{π} sin u d u = - π \cdot cos (u)]_{0}^{π} = - π (- 1 - 1) = 2 π$ .

Page last validated 2010-08-15