This module deals with various problems that can be modeled using integral calculus. As in the previous sections, the problem will be to find the total accumulation of some quantity , and the method will be to determine a slice of the quantity, the U element , and integrate.


Mass of a rod

Consider the problem of determining the mass of a rod. Suppose the rod's density varies along the length of the rod (but the rod is uniform in cross section). Let denote the linear density (i.e. the mass per unit of length) of the rod at position :

Then the mass element is the density times the thickness of the slice , as shown above, and it follows that the mass of the rod is

Mass of the earth

Consider the problem of finding the mass of the earth. Suppose the density of the earth is given as a function of the distance from the center of the earth. Assume that there are just three layers (inner core, outer core, and mantle) and that the density is constant within each layer.

What is the mass element in this case? It is important to note that in this example we are measuring the contribution of a spherical shell to the mass of the earth. This contribution is the volume of the spherical shell multiplied by the density of the shell. Mathematically,

Recalling that the surface area of a sphere of radius is , we have that the volume element is

and so the mass element is


Using the approximate graph of density above, estimate the mass of the earth.

Note that our volume is being measured in cubic kilometers, but the density is in grams per cubic centimeter. We need a conversion factor to make sure the units come out correctly. A little unit conversion gives us that

So we need to multiply by this so that the units are correct (and the final answer will be in kilograms).

Splitting the integral based on the values of for which is constant, we find

According to Wolfram Alpha, the mass of the earth is approximately kilograms, so our rough estimate is not too far off.


Imagine a rod of variable density which is attached to a hinge. The torque at the hinge depends not just on the weight of the rod but on the distribution of the weight.

If there were just a mass-less rod with a single point mass, the torque would be . This can be used to determine the torque element by thinking of each slice of the rod as a point mass. What is the torque on such a slice?

First, the torque element is the distance from the hinge, , times the force element (the force on the slice). The force element is the mass of the slice times the gravitational constant . Finally, as in the previous example, the mass element . Putting it all together, one finds

Integrating this over the length of the rod gives the torque.

Hydrostatic force

The next application is to compute the total force exerted by a tank of fluid on a surface submerged in the tank, often called the hydrostatic force. For a tangible example, consider a large aquarium with a circular glass viewing window (see the diagram below). If the viewing window has radius , and the top of the viewing window is at depth , then the problem is to find the total force of the water on the viewing window.

As always, the method will be to find the force element (the force on a small strip of the window), and then use integration to find the total force.

Recall that if pressure is constant across a surface, the force on the surface is . Hydrostatic pressure is given by

Note the units: , which is the correct unit for pressure (force per unit of area). Since the density of the fluid is assumed to be constant, the pressure only depends on the depth.Therefore, the most logical choice for the force element is a horizontal strip, since the depth, and hence the pressure, will be constant across the strip. Letting denote the area of the strip, we find that the force element is given by

where is the weight density of the fluid and is the depth of the strip.


Compute the total force exerted on the circular viewing window in the aquarium shown above.

As mentioned above, we will use horizontal strips of the window as the area element. The force element is the amount of force on that strip of the window. Let be the distance from the center of the window to the horizontal strip. Let up be negative, down be positive (so the top of the window is and the bottom of the window is :

Then the depth of the strip is , and the area of the strip is . Thus the force element in this example is

where is the weight density of water. So

Now, notice that is an odd function, so its integral from to is 0. Thus

since gives half the area of a circle of radius .

It is worth observing that with a very symmetric window such as the circle in this example, one can take the area of the window, , and multiply by the pressure at the center of the window , to find the hydrostatic force:

The reason this works is that the pressure on a horizontal strip above the center of the window averages with the pressure on the strip's mirror image below the center to give the pressure at the center of the window.


Compute the force on the endcap of a full cylindrical tank of radius on its side.

Using the knowledge gleaned from the previous example, we can take the area of the endcap, , and multiply by the hydrostatic pressure at the center of the endcap, which is , to find that the force is

We could also note that this is really a special case of the aquarium window example above, by setting in that example.


Consider a dam in the shape of a trapezoid with height , top edge and bottom edge . Find the total force exerted on the dam by the water:

As above, the force element is the force exerted on a horizontal strip. Let be the distance of the horizontal strip from the top of the dam, and be the length of the strip

Since the shape is a trapezoid, is a linear function of , and from the top and the bottom of the dam, one finds that and . It follows from the slope intercept form of a line that .

So the force acting on the strip is , where is the weight density of the water, is the depth of the strip, and is the area of the strip. Putting it all together, one finds

Present value

Consider the problem of determining the present value of some amount of money at a future time. Turning the problem around, first consider the value of an initial amount of money at a future time . Assuming a constant annual nominal interest rate and continuous compounding, this problem was an example of exponential growth, and had solution

where is the time in years. Given some amount of money, , at time , finding its present value is a matter of solving for . In other words, solving for present value in this simple case is the same as finding the initial investment which yields after years of continuous compounding interest. Solving this equation gives that the present value of a future amount at time is given by


Find the present value of $1000000 in 30 years, assuming an interest rate of .

From the above equation one finds that

Now consider an income stream, say from a job. If is the rate of income at time , what is the present value of that income stream? Let be the present value. Then consider the income earned over a small amount of time years in the future:

(the income element). This small bit of income at time contributes to the present value of the income stream. Thus the present value element is given by

Integrating this over the range of values of (the time period of the income stream) gives the present value of that income stream.


The Bigbucks lottery has an option of either a single lump sum payment today or an annuity which pays a constant amount each year for 20 years. Suppose the annuity pays $3 million a year (for 20 years), and that the interest rate will remain steady at . What is the fair lump sum payout today?

The income stream is constant at . Thus,

which is approximately $38 million.


  • Consider a dam with the shape of an isosceles triangle. The base of the triangle, which is parallel to the ground, is 5m long, and the height of the triangle is 10m. The weight density of water is given by . Compute the force exerted on the dam by water.