Triple Integrals

If is a function defining the mass density of an object:

then the integral is just the mass, and

Suppose the domain of integration can be described as (see set builder notation).

The integral of over the domain can be expressed as:

Evaluate

where is the volume bounded by the tetrahedron .

solution

$$ \int_{0}^1 \left(\int_{0}^{1 - y} \left[ \int_{0}^{1 - x - y} z \, dz \right] \, dx \right)\, dy = \frac{1}{24} $$

The domain of integration is always the "biggest" part of the shape. This is pretty obvious when you think about it.

Notice that our integrals evaluate to a number, because they follow this pattern:

Every time we integrate, we reduce the dimension by one. From the example above, for our first integral, we're integrating over a 3-D region on a 2-D domain. Next, we're integrating over a 2-D region on a 1-D domain. Finally, we're integrating over a 1-D region with a domain consisting of a point.

Evaluate where is the volume contained in the paraboloid and .

solution

The integrand has no , and there's lots of symmetry along the y-axis, which means there might be change of variable later, so we should integrate on first.

Next, we integrate on :

We don't use , and instead set because at the beginning, we already decided that we were "shooting our arrows" from (see diagram, green arrow).

Finally,

now we will apply change of variable to change to Polar Coordinates:

Taking the last example, we convert to Cylindrical Coordinates, except our y-axis is the "Cartesian part".

First, finding the Jacobian:

Since :

Mount Fuji is an approximately conical mountain 3800m tall and radius 19000m with a mass density of . How much work did it take to build Mount Fuji?

solution

Recall the formula for work:

we have:

so

We need our bounds

Taking the edge of the cone on the yz-plane:

Now, generalizing this for all :

Since we have for mass density, we're going to use and for our cylindrical coordinates.

Plugging this back into our integral:

Calculate the volume of bounded by .

solution

Simplifying the bounds:

Taking our equation , if we set , then we have . So our integration bounds look like:

Why?

• For , we're technically dependent on and , but doesn't matter since it's a circle. So our radius depends on the angle of the "arm", and using a right-angled triangle, we get .
• For , we're going all the way around the sphere, so we know right away it's
• For , we go from a flat line to a vertical line, which is , or

Now, using substitution, let: , and our new bounds are and .