The trough pictured below has length L and a cross-section defined by the curve y = a x^(3/2), up to height H. We wish to find the volume of the trough.

• We will proceed by considering vertical slices of the trough, taken over intervals of thickness `dyi at position yi.
• To find the volume of a slice, we wish to know the cross-sectional area of the trough, which will be the area of the z vs. x graph at the top of the figure below.
• We can find this area by determining the x interval corresponding to the z = a x^(3/2) curve and the height H.
• If we let z = H, we find as indicated that the width w of the region is 2 ( H / a ) ^ (2/3).

The cross-sectional area of the region will be the difference between the area under the curve and the area of the rectangle between the x axis and z = H.

The figure below depicts a side view of the trough with the interval `dyi over which we are attempting to find the volume.

• The calculation of the cross-sectional area is depicted below.

The volume will be the sum of all the volumes dVi = c.s. area * `dyi.

• Since the cross-sectional area is constant along the length of the trough, it can be factored out of the sum and we see that the volume is simply the cross-sectional area times the sum of the `dyi length increments.
• Since the length increments add up to the total length L, we see that the volume is equal to the product of the cross-sectional area and the length.

In the figure below we begin to determine the volume of a section of a cone which has been sliced perpendicular to its vertical axis of symmetry.

• The section's diameter decreases from 40 ft to 20 ft over a vertical distance of 15 feet.
• We will first calculate the 'local' volume of a horizontal slice of thickness `dyi at altitude yi; we will then sum these contributions and take a limit to get the total 'global' volume.

If we let `dri stand for the change in the radius between y = 0 and y = yi, then since the change in radius between y = 0 and y = 15' is 10', `dri / yi = 10 / 15 and `dri = 2/3 yi.

• It follows that the radius ri is 20 ft - `dri = 20 ft - 2/3 yi.
• The area of the cross-section is therefore Ai = `pi ri^2 = `pi (20 - 2/3 yi)^2 and the volume is Vi = Ai `dyi.
• The resulting integral for the volume, with y running from 0 to 15, is depicted below.

If the density of the contents of a vertical tower 40 cm high range linearly with altitude from 3 g / cm^3 at the top to 4 g / cm^3 at the bottom of the tower, then if the tower has cross-sectional area 100 cm^2, what is the mass of the tower's contents?

• We consider a horizontal strip of thickness `dxi at altitude xi.
• The density at position xi is 4 - .025 xi (density is a linear function with value 4 at x = 0 and 3 at x = 40; slope and intercept of the linear function are obvious).

The mass contained in the strip is equal to the product of the volume of the strip and the density at the altitude of the strip.

• The volume of the strip is Vi = c.s. area * `dxi.
• The mass of the ith strip is therefore as indicated below.

We write a Riemann sum for the total mass, using the local expression obtained above, and take the limit to obtain the indicated integral, where altitude x runs from 0 to 40.

• The resulting total mass is 14000 grams.

A cylinder with cross-sectional area A and height 10,000, with density function `rho(y), has a total mass that we determine by first considering the horizontal slice of thickness `dyi at altitude yi.

• The slice is a cylinder with base area A and altitude `dyi, so its volume is Vi = A `dyi.
• The mass of the slice is therefore mi = `rho(yi) Vi = `rho(yi) A `dyi.

We wish to find the horizontal line on which the object in the figure below will balance.

• Having cut the actual object out of uniform cardboard, we will check our results by actually balancing the object at the position we determine.
• We let x stand for the distance from the 15 cm edge, and consider the mass of a horizontal slice of thickness `dxi at x = xi.
• We assume a constant density `sigma, which is an area density (e.g., measured in grams/cm^2).
• If the area of our slice is Ai, then its mass is mi = `sigma * Ai.
• The length of the slice is found from similar triangles as indicated below to be 15 cm - 9/31 xi, so its area is Ai = `sigma * `dxi.

If we were to stack the slices at appropriate distances on a support of negligible mass, then the slices would exert a torque `tau about x = 0.

• This torque will be the same as the torque exerted about x = 0 by the total mass at the position of the center of mass.

The torque exerted by any mass about x = 0 is considered here to be the product of the mass and its distance (called the 'moment arm') from the axis of rotation (i.e., from x = 0).

• Present and future physics students note that torque is actually defined in physics to be the product of force and moment arm; here we are using mass instead of force, which is technically incorrect. However, for the purposes of finding the center of mass, we can get away with this, since the force will be proportional to the mass: we would find a center of mass by dividing torque by weight, and it would turn out that gravitational acceleration g would cancel and we would get the same result we will obtain here.

We carry out the details below.