Triple Integral in Cartesian Coordinates

TRIPLE INTEGRAL IN CARTESIAN COORDINATES

Let f(x, y, z) be a continuous function at every point in a closed and bounded region D in space. Subdivide the region into a number of element volumes by drawing planes parallel to the coordinate planes. Let ΔV1, ΔV2, …., ΔVn be the number of element volumes formed. Let (x1, y1, z1) be any point in ΔV1, where 

As in the case of double integrals, the triple integral is evaluated by three successive integration of single variable.

Consider the triple integral

(1) If all the limits are constants, then the integration can be performed in any order with proper limits,

First we integrate w.r.to x, treating y and z as constants and substitute limits of x. Next integrate the resulting function of y and z w.r.to y, treating z as constant and substitute the limits of y. Finally we integrate the resulting function of z w.r.to z and substitute the limits of z.

WORKED EXAMPLES

Example 1

Evaluate

Solution 

Example 2 

Evaluate

Solution 

Example 3 

Evaluate

Solution 

Example 4 

Evaluate

Solution 

Example 5 

Evaluate

Solution 

Example 6 

Evaluate

Solution 

Example 7 

Evaluate

Solution 

Example 8 

Evaluate

Solution 

Example 9 

Evaluate

Solution 

Example 10 

Evaluate

Solution 

Example 11 

Evaluate

Solution 

Example 12 

Evaluate

Solution 

Example 12(a) 

Evaluate

Solution 

Example 13 

Evaluate

Solution 

Example 14 

Evaluate over the volume V enclosed by the three coordinate planes and the plane

Solution 

Let V be the volume enclosed by the plane the coordinate axes in A(a, 0, 0), B(0, b, 0), C(0, 0, c) respectively. 

The projection of V on the xy-plane is the ΔOAB bounded by

1. Change of variables in triple integral

The evaluation of a triple integral, sometimes become simpler if the variables of integration are transformed suitably into new variables.

1. Change of variables from x, y, z to the variables u, v, w.

Let   be the given triple integral

Usually we choose cylindrical polar coordinates or spherical polar coordinates.

2. Change the variables to cylindrical polar coordinates (p, Ø, z)

Let be the triple integral. Changing the Cartesian coordinates (x, y, z) to cylindrical polar coordinates (p, q, z), we have x = pcos φ, y = psin φ, z = z and f(x, y, z) will be F(p, q, z).

The Jacobian of transformation is

3. Change the variables to spherical polar coordinates (r, θ, φ)

Let be the given triple integral changing to the carterian coordinates (x, y, z) to spherical polar coordinates (r, θ, φ), we have x = r sin θ cos φ, y = r sin θ sin φ, z = rcos θ and ƒ(x, y, z) will be F(r, θ, φ) 

The Jacobian of transformation is

2. Volume as triple integral

Triple integrals can be used to evaluate volume V of a finite bounded region D in space.

The volume V =

 [Taking f(x, y, z) = 1 in (1) of 5.2 page 5.64, we get volume]

WORKED EXAMPLES

Example 1 

Change to spherical polar coordinates and hence evaluate where V is the volume of the sphere x2 + y2 + z2 = a2.

Solution 

Example 2 

Find the volume of the region bounded by the paraboloid z = x2 + y2 and the plane z = 4.

Solution: 

The volume of the region bounded by z = x2 + y2 and z = 4 is

The section of the paraboloid

z = x2 + y2 by the plane z = 4 is a circle x2 + y2 = 4

The projection of the region in the xy plane will be the area of the circle x2 + y2 = 4.

From Fig. 5.58 we see z varies from x2 + y2 to 4.

Put x = 2 sinθ

⸫ dx = 2cosθdθ

Example 3 

Find the volume of the region of the sphere x2 + y2 + z2 = a2 lying inside the cylinder x2 + y2 = ay.

Solution 

x2 + y2 = ay 

x2 + y2 - ay = 0

So, the cylinder has this circle as guiding curve and generators parallel to z-axis.

x2 + y2 + z2 = a2 is a sphere with centre (0, 0, 0) and radius = a.

The volume inside the cylinder bounded by the sphere is symmetric about xy-plane. So the required volume = 2 (volume inside the cylinder) above the xy-plane.

Its projection in the xy-plane is the circle x2 + y2 = ay. The circle is symmetric about y-axis.

where D is the common region in the first octant.

Changing to cylindrical polar coordinates (r, θ, z), we have x = rcos θ, y = rsin θ, z = z

Example 4 

Find the volume cut off from the sphere x2 + y2 + z2 = a2 by the cone x2 + y2 = z2.

Solution 

The cone x2 + y2 = z2 is a circular cone with vertex origin and semivertical angle 

The volume cut off by the cone from sphere is equal in all the 8 octants.

⸫ Required volume = 8 (volume bounded in the I octant).

where D is the region in the I octant.

The projection of D in the xy-plane is the region bounded by the circle x2 + y2 = a2. 

We change cartesian coordinates

Example 5 

Find the volume of the cylinder x2 + y2 = 4 bounded by the plane z = 0 and the surface z = x2 + y2 + 2.

Solution 

The region is bounded by the cylinder x2 + y2 = 4 above the xy-plane and the surface z = x2 + y2 + 2.

Changing to cylindrical polar coordinates, we get

Example 6 

A Circular hole of radius b is made centrally through a sphere of radius a. Find the volume of the remaining sphere.

Solution 

Both the sphere and circular hole are symmetric about the xy plane. So volume of the hole = 2 × volume of the hole above the xy-plane

V is the volume above the xy-plane

Where the region R is the circle

x2 + y2 = b2, b is the radius of hole and x, y vary over R.

By changing to polar coordinates, we shall evaluate this double integral.

⸫ Put x = r cos θ,  y = r sin θ,

⸫ dx dy = r dr dθ, x2 + y2 = r2

r varies from 0 to b and θ varies from 0 to 2π

Example 7 

Evaluate over the positive octant of the sphere x2 + y2 + z2 = a2 by transforming into spherical coordinates

Solution: 

Let I = where R is the region of the first octant bounded by the sphere x2 + y2 + z2 = a2. Changing the carterian coordinates (x, y, z) to the spherical polar coordinates (r, θ, ϕ), we have x = rsin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.

Example 8 

Evaluate 

Solution 

Changing the carterian coordinates (x, y, z) to the spherical polar coordinates (r, θ, ϕ), we have

x = rsin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.

Example 9 

Find the volume of the tetrahedron bounded by the plane and the coordinate planes.

Solution 

The region of integration is the region bounded by

Its projection in the xy-plane is the ΔOAB bounded by x = 0, y = 0,

Example 10 

Find the volume of sphere x2 + y2 + z2 = a2 using triple integrals.

Solution 

Since the sphere x2 + y2 + z2 = a2 is symmetric about the coordinate planes, the volume of the sphere

Example 11 

Find the volume bounded by the cylinder x2 + y2 = 4 and the planes y + z = 4, z = 0.

Solution 

Required volume of the cylinder x2 + y2 = 4, cut off between the planes

Example 12 

Find the volume common to the cylinders x2 + y2 = a2and x2 + z2 = a2.

Solution 

The cylinder x2 + y2 = a2 has its generators parallel to z-axis, axis is z-axis and base circle x2 + y2 = a2 is in the z = 0 plane.

The cylinder x2 + y2 = a2 has its generators parallel to y-axis, axis is y-axis and base circle x2 + y2 = a2 is in y = 0 plane. Both the cylinders have common volume equally in all 8 octants. So the required volume V = where D is the common region in the I octant

Example 13 

Find the volume of the ellipsoid

Solution 

Note: If a = b = c, the ellipsoid becomes the sphere x2 + y2 + z2 = a2. The volume of the sphere.

Example 14 

Evalute where V is the region bounded by x = 0, y = 0, z = 0 and x + y + z = 1.

Solution

Given x + y + z = 1,

⸫ z = 1 - x - y

⸫  z - varies from z = 0 to z = 1 - x - y

When z = 0, x + y = 1 ⇒ y = 1 - x

⸫ y - varies from y = 0, y = 1 - x

When y = 0, z = 0, x = 1

⸫ x varies from x = 0, to x = 1

EXERCISE 

Volume as Triple Integral

17. Find the volume in the positive octant bounded by the plane

x + 2y + 3z = 4 and the coordinate planes.

ANSWERS TO EXERCISE