Archive for the ‘Trigonometric Substitution’ Category

Pressure is a force per unit of area exerted over the surface of an object (as in 35 psi in the tires on your car). When an object is immersed in water, or some other liquid medium, the fluid pressure exerted on that object varies with the depth at which it is submerged. For example, the volume of water pressing down on an object submerged 10 feet is twice that of an object submerged half that depth.

Fluid Pressure (force per unit area) can therefore be defined as follows:

p=wh ,where w = weight density of the fluid h = depth at which the object is submerged

Fluid Force (total force exerted on object) is therefore given by

F=pA ,where A=total area of surface object in question F=whA

According to Pascal’s Law (principle), an object submerged in a fluid is subjected to equal pressure in all directions (at any given depth). For a sheet of metal submerged in water and resting horizontally at a given depth, the fluid force is constant over its entire surface. If, however, the submerged sheet is resting vertically, the entire force exerted over this object by the water varies with depth; the bottom of the sheet will experience more fluid force than its top. In order to determine the total force acting on this vertically oriented sheet, the force exerted on each rectangular cross-section of infinitesimal width (Δy) will be calculated and summed over the object’s entire vertical span; enter integration. With a well-chosen location for the y-axis in our model, the length of each rectangular cross-section can be easily expressed as some variation of f(y).


Rectangular Plate

Fluid ForceRectangularPlate


The example above was relatively simple since f(y) was a constant throughout its vertical span. With shapes whose widths are not constant, the mathematical model can once again vary depending on the perspective chosen. In the examples below, circular plates have been introduced since they provide opportunities for multiple forms of substitution in the integration process, thereby maximizing learning opportunities for students.


The fluid force acting on the ends of a cylindrical water tank is the subject below. I felt that this would be more interesting than imagining the force exerted on a circular plate submerged in a body of water. To simplify matters here, atmospheric pressure and other factors such as sliding forces have been ignored.

Cylindrical Tank: Half-full (Perspective 1)



Cylindrical Tank: Half-full (Perspective 2)



Cylindrical Tank: Filled to Capacity



It is worthwhile drawing attention to the forces acting on the ends of top half of the tank filled to capacity (451.34 lb) and the bottom half filled to half capacity (332.8 lb). This type of thoughtful comparison can add to the students’ understanding of this topic.


The two images directly below once again show the two perspectives of fluid forces acting on the ends of our water tank. They are included here to illustrate the contents of the interactive link that follows.

Tank Centered at (0,0)



Tank Centered at (0,-2)


Click on the link provided here to interact with fluid force on the ends of a cylindrical tank.


In the image and link below, the circular end has been removed from the tank and submerged on its own. Once again, various mathematical models could be used to describe this scenario; the one chosen here has placed the origin at the circle’s center.


Click on the link provided here to interact with fluid force on a submerged vertical plate with center at origin.


The following links will be of interest to some:

Pressure and Buoyancy and Hydrodynamics


Thanks for reading.


Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath.

As mentioned earlier, my students’ first exposure to calculus is from the “Leibniz” perspective. Introducing integration in the first two weeks opens the door to the exploration of many interesting scenarios, including separable differential equations. These lead very nicely to discussions on rates of change,  providing a seamless segue into the land of differential calculus and derivatives.

After students have a firm hold on the concept of differential calculus, I like to shake things up once in a while by throwing in some integration problems. The subject of this entry, shown below, is one such problem. Students are asked to determine the area bounded by the function  f(x)=4/(1+x^2) and the x-axis from  x=0  to  x=1. When the result emerges, students are highly motivated to understand what’s going on. A great discussion ensues and a plan of attack is formulated; this first attempt is a good one but it eventually ends in a stalemate as illustrated below.


Why u-substitution fails

Pi Estimate Natural Log Attempt


Allowing students to travel down the wrong path often leads to greater learning than might otherwise occur; the example above is no exception. Learning various methods of integration is one thing, but knowing when to employ one approach over another is very empowering as well.

The impasse that was reached above provides motivation to seek out another approach; enter trigonometric substitution. This is a very powerful tool in the arsenal once students learn to recognize the circumstances in which it can be used. It is demonstrated below and leads to the conclusion of this problem.


Trigonometric Substitution

Pi Estimate Trig Sub

Other examples of trigonometric substitution can be seen in calculating  circumference  and area of the circle.

Later on this semester, students will be challenged to determined the arc length on the function  f(x)=x^2  over a given interval; trigonometric substitution will once again be deployed. This time, however, the solution will not be determined so easily. The roadblock in this scenario will lead to even greater learning, hyperbolic trig-substitution included.


Thanks for reading.

Surface areas of curved 3-dimensional solids tend to be much more difficult for students to conceptualize than those whose sides do not stray from a “level” plane. These will eventually be addressed but we will first discover how to calculate lengths of curves.

The circle will once again be called upon to initiate this exploration; the image below illustrates, in part, the method of exhaustion that Archimedes utilized to arrive at his estimate for π.

Archimedes Circumference2

Click on the link here to interact with what Archimedes revealed.


……..and now this. Was Archimedes wrong???



If the fellow above had joined pairs of points at each successive corner with a line segment (hypotenuse) and based his calculation for circumference on the sum of those, he would have found that Archimedes was correct all along.


Arc Length Formula

As  Δs approaches zero in this exploration, its length becomes a more accurate estimate for the arc length near the “point” of tangency (there are always two points in very close proximity). The end result through the limiting process shown directly below is the formula for arc length.

ArcLength Circle

Its always beneficial for students to work through several examples to cement their understanding of new concepts and related procedures; my preference is to provide examples that are already familiar to them. Calculating arc length (in this case) can then serve as a verification and acceptance of the new concept is achieved with confidence. The offering directly below and the link that follows connects the arc length formula to the Pythagorean theorem.

Arc Length

The formula for arc length is based on the Pythagorean theorem; it is therefore not surprising that they produce the same lengths on linear functions.


The real power of the formula for arc length lies in its applications to curves. Since students have known the circle’s circumference for several years, it is appropriate to now derive 2πr using our new tool. This is shown below and once again brings trigonometric substitution into play.

Circumference of the Circle

Arc Length (Circumference of Circle)

Once the circle’s circumference has been established using the arc length formula, the integration process can be further solidified by using arc length to once again calculate the circle’s area.


The formula for arc length will once again be employed in deriving the formulas for surface area of the sphere and cone.


Thanks for reading.


When working with composite functions, u-substitution often simplifies the integration process for us. What is sometimes overlooked, however, is the significant adjustment that occurs to the area beneath the curve.  The image and link below allow students to explore the significance of this procedure.

ChainRule Area Cos

Interactively explore u-substitution.


As seen below, u-substitution is not a viable approach to integration as its leads to division by zero. In the case below, and others like it, trigonometric substitution provides a means for us to move forward and solve many problems. The process below also brings trigonometric identities into play, thereby  exposing students to their significance.

Circle Area Trig Substitution Formula1

Circle Area Trig Substitution Formula2

Thanks for reading.