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Calculus III - Notes
 Applications of Partial Derivatives Previous Chapter Next Chapter Line Integrals Triple Integrals in Spherical Coordinates Previous Section Next Section Surface Area

Change of Variables

Back in Calculus I we had the substitution rule that told us that,

In essence this is taking an integral in terms of x’s and changing it into terms of u’s.  We want to do something similar for double and triple integrals.  In fact we’ve already done this to a certain extent when we converted double integrals to polar coordinates and when we converted triple integrals to cylindrical or spherical coordinates.  The main difference is that we didn’t actually go through the details of where the formulas came from.  If you recall, in each of those cases we commented that we would justify the formulas for dA and dV eventually.  Now is the time to do that justification.

While often the reason for changing variables is to get us an integral that we can do with the new variables, another reason for changing variables is to convert the region into a nicer region to work with.  When we were converting the polar, cylindrical or spherical coordinates we didn’t worry about this change since it was easy enough to determine the new limits based on the given region.  That is not always the case however.  So, before we move into changing variables with multiple integrals we first need to see how the region may change with a change of variables.

First we need a little notation out of the way.  We call the equations that define the change of variables a transformation.  Also we will typically start out with a region, R, in xy-coordinates and transform it into a region in uv-coordinates.

 Example 1  Determine the new region that we get by applying the given transformation to the region R. (a) R is the ellipse  and the transformation is , .   [Solution] (b) R is the region bounded by , , and  and the                   transformation is , .   [Solution]   Solution (a) R is the ellipse  and the transformation is , .   There really isn’t too much to do with this one other than to plug the transformation into the equation for the ellipse and see what we get.                                                                So, we started out with an ellipse and after the transformation we had a disk of radius 2.   (b) R is the region bounded by , , and  and the transformation is , .   As with the first part we’ll need to plug the transformation into the equation, however, in this case we will need to do it three times, once for each equation.  Before we do that let’s sketch the graph of the region and see what we’ve got.     So, we have a triangle.  Now, let’s go through the transformation.  We will apply the transformation to each edge of the triangle and see where we get.   Let’s do  first.  Plugging in the transformation gives,                                                       The first boundary transforms very nicely into a much simpler equation.   Now let’s take a look at ,                                                         Again, a much nicer equation that what we started with.   Finally, let’s transform .                                                     So, again, we got a somewhat simpler equation, although not quite as nice as the first two.   Let’s take a look at the new region that we get under the transformation.   We still get a triangle, but a much nicer one.

Note that we can’t always expect to transform a specific type of region (a triangle for example) into the same kind of region.  It is completely possible to have a triangle transform into a region in which each of the edges are curved and in no way resembles a triangle.

Notice that in each of the above examples we took a two dimensional region that would have been somewhat difficult to integrate over and converted it into a region that would be much nicer in integrate over.  As we noted at the start of this set of examples, that is often one of the points behind the transformation.  In addition to converting the integrand into something simpler it will often also transform the region into one that is much easier to deal with.

Now that we’ve seen a couple of examples of transforming regions we need to now talk about how we actually do change of variables in the integral.  We will start with double integrals.  In order to change variables in a double integral we will need the Jacobian of the transformation.  Here is the definition of the Jacobian.

Definition

 The Jacobian of the transformation ,  is

The Jacobian is defined as a determinant of a 2x2 matrix, if you are unfamiliar with this that is okay.  Here is how to compute the determinant.

Therefore, another formula for the determinant is,

Now that we have the Jacobian out of the way we can give the formula for change of variables for a double integral.

Change of Variables for a Double Integral

 Suppose that we want to integrate  over the region R.  Under the transformation ,  the region becomes S and the integral becomes,

Note that we used du dv instead of dA in the integral to make it clear that we are now integrating with respect to u and v.  Also note that we are taking the absolute value of the Jacobian.

If we look just at the differentials in the above formula we can also say that

 Example 2  Show that when changing to polar coordinates we have     Solution So, what we are doing here is justifying the formula that we used back when we were integrating with respect to polar coordinates.  All that we need to do is use the formula above for dA.   The transformation here is the standard conversion formulas,                                                   The Jacobian for this transformation is,                                                     We then get,

So, the formula we used in the section on polar integrals was correct.

Now, let’s do a couple of integrals.

 Example 3  Evaluate  where R is the trapezoidal region with vertices given by , ,  and  using the transformation  and .   Solution First, let’s sketch the region R and determine equations for each of the sides.   Each of the equations was found by using the fact that we know two points on each line (i.e. the two vertices that form the edge).    While we could do this integral in terms of x and y it would involve two integrals and so would be some work.   Let’s use the transformation and see what we get.  We’ll do this by plugging the transformation into each of the equations above.   Let’s start the process off with .                                                                Transforming  is similar.                                                             Next we’ll transform .                                                          Finally, let’s transform .                                                           The region S is then a rectangle whose sides are given by , ,  and  and so the ranges of u and v are,                                                    Next, we need the Jacobian.                                                    The integral is then,

 Example 4  Evaluate  where R is the ellipse given by  and using the transformation , . Solution The first thing to do is to plug the transformation into the equation for the ellipse to see what the region transforms into.                       Or, upon dividing by 2 we see that the equation describing R transforms into                                                                   or the unit circle.  Again, this will be much easier to integrate over than the original region.   Note as well that we’ve shown that the function that we’re integrating is                                                        in terms of u and v so we won’t have to redo that work when the time to do the integral comes around.   Finally, we need to find the Jacobian.                                              The integral is then,                                          Before proceeding a word of caution is in order.  Do not make the mistake of substituting  or  in for the integrands.  These equations are only valid on the boundary of the region and we are looking at all the points interior to the boundary as well and for those points neither of these equations will be true!   At this point we’ll note that this integral will be much easier in terms of polar coordinates and so to finish the integral out will convert to polar coordinates.

Let’s now briefly look at triple integrals.  In this case we will again start with a region R and use the transformation , , and  to transform the region into the new region S.  To do the integral we will need a Jacobian, just as we did with double integrals.  Here is the definition of the Jacobian for this kind of transformation.

In this case the Jacobian is defined in terms of the determinant of a 3x3 matrix.  We saw how to evaluate these when we looked at cross products back in Calculus II.  If you need a refresher on how to compute them you should go back and review that section.

The integral under this transformation is,

As with double integrals we can look at just the differentials and note that we must have

We’re not going to do any integrals here, but let’s verify the formula for dV for spherical coordinates.

 Example 5  Verify that  when using spherical coordinates.   Solution Here the transformation is just the standard conversion formulas.                                The Jacobian is,                 Finally, dV becomes,                                        Recall that we restricted  to the range  for spherical coordinates and so we know that  and so we don’t need the absolute value bars on the sine.

We will leave it to you to check the formula for dV for cylindrical coordinates if you’d like to.  It is a much easier formula to check.

 Triple Integrals in Spherical Coordinates Previous Section Next Section Surface Area Applications of Partial Derivatives Previous Chapter Next Chapter Line Integrals

[Notes]

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