# 9. Double and multiple integrals, line integrals

### Double integrals

Given a function and a set , the integral

is simply the volume under the graph of over the given set.

Assume that we would like to calculate the volume of a pyramid with a base in form of a isosceles right triangle (with boundary of axes and line ) and height , so in point the height is given by the following formula . So the volume is

where is the triangle mentioned above.

How to calculate this integral? It will be described below.

### Fubini Theorem

Fubini theorem states that such an integral is the integral over of the integral over , and i the same as integral over of the integral over , in other words (for functions nice enough):

where and are such that

Obviously you have to treat the outer variable as a constant parameter when calculating the inner integral.

Therefore, returning to our example of the pyramid, the section at has area , so the volume is:

which is consistent with our knowledge about geometry.

### Integration with change of variables

If is a diffeomorphism, the following theorem holds

,
where , where .

### Polar coordinates

The most common use of the above method are polar coordinates which are defined by the diffeomorphism . Thus, .
Moreover,

so

E.g. let us calculate

We get:

### Volume

If we want to calculate the volume of a set we can do it using one of the following three methods:

where is the basis is the height at and is the area of the section for given .

Assume that we would like to calculate the volume of a pyramid with a base in form of a isosceles right triangle (with boundary of axes and line ) and height , so in point the height is given by the following formula . Therefore the section at has area , so the volume is:

which is consistent with our knowledge about geometry.

### Volume of a solid of revolution

If we take a graph of a function and revolve it around axis , on the interval , we will get a solid of revolution. Since the area of a section for given equals , from the above considerations we get that the volume equals .

### Integrals in natural sciences

Very often physical variables can be described as a surface area or volume under a graph of another variable, i.e. as an integral. E.g. the change of position is the area below the graph of speed (because speed is the derivative of position), the force applied by a fluid is the integral over pressure, the mass of an object is an integral over its density at each point. Such situations can be expected whenever if the variable is constant on the whole length/ area / volume we would multiply it by length / area / volume.

### Average value of a function

Similarly the average value of a function (i.e. average height of its graph), is simply the area/ volume under the graph (i.e. the integral) divided by the length/area of the base.

### Centre of mass

The centre of mass along axis, is a point , such that the torque of weight (i.e. length of lever arm times mass times a constant, which here can be omitted) is equalized. Let us say that the mass of a section for given is . The lever arm length . So we get that , where and are ends of a studied object. Thus, , where is the whole mass. Thus, . Notice that in order to calculate and we usually also have to calculate integrals.

### Line integral of a scalar field

Imagine a curve described as a parametrization (e.g. as a route of a moving point). Assume that there is a value associated with each point, e.g. density of the material. We can calculate the mass of the whole curve between and . It is clear that since the change of position is the are between the graph of speed and the velocity is and we get speed , so the length of the curve is . Thus, to calculate its mass we have to multiply it in every point by the density, i.e.

and this is actually what is called the line integral of a scalar field.

### Surface are of a solid of revolution

Thus, if we have a function , its graph is described by the parametrization , . Thus, its length is simply

Now, if we revolve it around axis, then at given we get circumference , so the surface area of the solid of revolution is

### Line integral of a vector field

Let us imagine (imagine a rigid wire), described by a parametrization , on an interval . Imagine a bead moving along this wire. This time, not a scalar but a vector is associated with every point of the plane, e.g. a force which would be applied on the bead in this point. Let this force have an -coordinate and -coordinate .

We would like to calculate the work which is subtracted (or added) by this forces when moving the bead in the indicated way along the wire. This is exactly the line integral of the vector field and is denoted by:

and equals

which is nicely shown by this animation (by Wikipedia).

### Line integration over closed curve

If the curve is closed it is important to set its orientation (we would always assume anti-clockwise orientation, in the clockwise case one has to multiply the integral by ). We write then:

Green Theorem allows us to change such an integral into double integral, namely

where is the area cut out by the curve .