How To Find Change Of Coordinates Matrix

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Section four-8 : Modify of Variables

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

\[\int_{{\,a}}^{{\,b}}{{f\left( {g\left( x \right)} \correct)\,g'\left( ten \correct)\,dx}} = \int_{{\,c}}^{{\,d}}{{f\left( u \right)\,du}}\hspace{0.25in}{\mbox{where }}u = yard\left( x \right)\]

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 washed this to a sure extent when nosotros converted double integrals to polar coordinates and when we converted triple integrals to cylindrical or spherical coordinates. The principal deviation is that we didn't actually go through the details of where the formulas came from. If you recall, in each of those cases nosotros commented that we would justify the formulas for $dA$ and $dV$ eventually. Now is the time to do that justification.

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

First, nosotros need a little terminology/notation out of the style. We call the equations that define the change of variables a transformation. Also, nosotros volition typically start out with a region, $R$, in $xy$-coordinates and transform it into a region in $uv$-coordinates.

Case 1 Decide the new region that nosotros go by applying the given transformation to the region $R$.

$R$ is the ellipse ${10^two} + \frac{{{y^two}}}{{36}} = 1$ and the transformation is $x = \frac{u}{ii}$, $y = 3v$.
$R$ is the region divisional past $y = - ten + four$, $y = x + 1$, and $\displaystyle y = \frac{x}{3} - \frac{four}{3}$ and the transformation is $\displaystyle x = \frac{i}{2}\left( {u + v} \right)$, $\displaystyle y = \frac{ane}{2}\left( {u - v} \correct)$.

Prove All SolutionsHide All Solutions

a $R$ is the ellipse ${10^2} + \frac{{{y^two}}}{{36}} = 1$ and the transformation is $10 = \frac{u}{two}$, $y = 3v$. Show Solution

There really isn't likewise much to practice with this one other than to plug the transformation into the equation for the ellipse and see what nosotros get.

\[\begin{align*}{\left( {\frac{u}{ii}} \right)^2} + \frac{{{{\left( {3v} \right)}^2}}}{{36}} & = 1\\ \frac{{{u^ii}}}{iv} + \frac{{9{five^2}}}{{36}} & = 1\\ {u^2} + {five^2} & = 4\end{align*}\]

So, we started out with an ellipse and subsequently the transformation we had a disk of radius 2.

b $R$ is the region divisional by $y = - x + four$, $y = x + ane$, and $\displaystyle y = \frac{x}{3} - \frac{four}{3}$ and the transformation is $\displaystyle x = \frac{1}{two}\left( {u + v} \right)$, $\displaystyle y = \frac{1}{2}\left( {u - five} \right)$. Evidence Solution

As with the first part we'll need to plug the transformation into the equation, however, in this case we volition need to exercise it three times, once for each equation. Before nosotros exercise that allow'due south sketch the graph of the region and encounter what we've got.

$This is the 2D graph on an xy axis sytem of a triangle with vertices (4,0), $\left(\frac{3}{2}, \frac{5}{2} \right)$ and $\left(-\frac{7}{2}, -\frac{5}{2} \right)$. The top right edge of the triangle is given by $y=-x+4$. The top left edge is given by $y=x+1$. The bottom edge is given by $y=\frac{1}{3}x-\frac{4}{3}$. The triangle has been shaded in.$

So, nosotros have a triangle. Now, let'southward go through the transformation. We will apply the transformation to each edge of the triangle and see where we go.

Let'southward do $y = - x + 4$ first. Plugging in the transformation gives,

\[\begin{align*}\frac{1}{2}\left( {u - 5} \correct) & = - \frac{1}{2}\left( {u + v} \right) + 4\\ u - v & = - u - v + viii\\ 2u & = 8\\ u & = 4\end{align*}\]

The first boundary transforms very nicely into a much simpler equation.

Now let'south take a look at $y = x + 1$,

\[\begin{marshal*}\frac{ane}{2}\left( {u - five} \correct) & = \frac{ane}{2}\left( {u + five} \right) + 1\\ u - v & = u + v + 2\\ - 2v & = ii\\ five & = - 1\end{align*}\]

Again, a much nicer equation that what we started with.

Finally, let's transform $y = \frac{x}{three} - \frac{4}{iii}$.

\[\begin{marshal*}\frac{1}{2}\left( {u - five} \right) & = \frac{1}{3}\left( {\frac{i}{two}\left( {u + v} \right)} \right) - \frac{iv}{3}\\ 3u - 3v & = u + five - viii\\ 4v & = 2u + 8\\ 5 & = \frac{u}{ii} + 2\terminate{align*}\]

And then, again, nosotros got a somewhat simpler equation, although not quite every bit nice as the first ii.

Permit'southward take a look at the new region that we get under the transformation.

$This is the 2D graph on a uv axis system of a triangle with vertices (4,4), (4,-1) and (-6,-1). The top edge of the triangle is given by $v=\frac{1}{2}u+2$. The right edge is given by $u=4$. The bottom edge is given by $v=-1$. The triangle has been shaded in.$

We however get a triangle, but a much nicer i.

Annotation that we tin't always expect to transform a specific type of region (a triangle for example) into the same kind of region. Information technology is completely possible to have a triangle transform into a region in which each of the edges are curved and in no mode 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 beginning of this fix of examples, that is ofttimes 1 of the points behind the transformation. In improver to converting the integrand into something simpler it will often as well transform the region into one that is much easier to bargain with.

Before proceeding with the next topic let'due south accost another point. On occasion, nosotros will also need to know the range of $u$ and/or $v$ for each of the new equations we get from the transformation. Nosotros didn't demand that for the two examples higher up and it is not something that we volition often need. Still, information technology can help on occasion in determining the new region.

And then, permit'south work a quick example to see how we do that.

Example 2 For the region bounded by $y = - 10 + iv$, $y = x + 1$, and $y = \frac{x}{3} - \frac{4}{3}$ and the transformation $x = \frac{1}{2}\left( {u + v} \right)$, $y = \frac{1}{2}\left( {u - v} \right)$ determine the ranges of $u$ and $five$ for each of the new equations from the transformation.

Show Solution

Okay, nosotros already know what the new region looks like and what the new equations are from the previous case. Then, here is a quick review of the transformation of each of the original equations.

\[\brainstorm{marshal*} & y = - x + iv& \Rightarrow & \hspace{0.25in}u = 4\\ & y = x + i& \Rightarrow & \hspace{0.25in}5 = - 1\\ & y = \frac{x}{iii} - \frac{4}{iii}& \Rightarrow & \hspace{0.25in}v = \frac{u}{2} + 2\end{marshal*}\]

Hither is the new region we get nether the transformation.

Note, that, in this case, we could determine the range of $u$ and $5$ for each equation from the sketch above. However, in cases where we might actually need the ranges that is usually non an option as nosotros often need the ranges for $u$ and/or $v$ to become an accurate sketch of the new region.

And then, permit's now really commencement working the trouble.

Let's get-go with the equation $u = four$. Starting time, nosotros don't need a "range" of $u$'south hither as equation makes it pretty articulate we have a unmarried value of $u$, namely $u = 4$. So, allow'southward determine the range of $five$'s nosotros should get.

Allow's beginning with the $x$ transformation and plug in the known value of $u$ for this equation. That gives,

\[x = \frac{1}{two}\left( {u + v} \correct) = \frac{one}{2}\left( {4 + v} \right)\]

Now, we know that the range of $x$'s for the original equation, $y = - 10 + 4$, are $\frac{3}{2} \le 10 \le 4$. We as well know from in a higher place what $x$ is in terms of $five$, then plug that into this range and practice a footling manipulation as follows,

\[\begin{array}{c}\displaystyle \frac{three}{2} \le x \le 4\\\displaystyle \frac{3}{ii} \le \frac{ane}{2}\left( {4 + v} \right) \le 4\\ iii \le iv + five \le eight\\ - 1 \le v \le 4\finish{array}\]

So, the range of $5$'s for $u = iv$ must be $ - ane \le v \le 4$, which nicely matches with what we would expect from the graph of the new region.

Note that we could but as easily used the $y$ transformation and $y$ range for the original equation and gotten the same effect.

Okay, let's now motility onto $5 = - 1$ and we won't put in quite as much explanation for this part.

First, nosotros don't need a range of $v$ for this because we conspicuously accept just a single value of $five$. So, to get the range of $u$ allow's again beginning with the $ten$ transformation, plug $v = - 1$ in that and then use the range of $x$'due south from the original equation, $y = x + one$.

Here is that work.

\[\begin{assortment}{c}\displaystyle - \frac{7}{two} \le x \le \frac{3}{two}\\ \displaystyle - \frac{vii}{2} \le \frac{ane}{2}\left( {u - one} \right) \le \frac{3}{2}\\ - 7 \le u - 1 \le 3\\ - vi \le u \le four\terminate{array}\]

And so, the range of $u$ for $v = - one$ is $ - 6 \le u \le 4$ which, again, matches upward with what we come across on the graph. Besides annotation that once once again, nosotros could have used the $y$ ranges to practise this piece of work.

Finally, let's observe the range of $u$ and $v$ for $5 = \frac{u}{2} + 2$. This time allow's use the $y$ transformation so we can say nosotros used that in one these. Then, we'll commencement with the range of $y$'s for the original equation, $y = \frac{x}{3} - \frac{4}{3}$, plug in the $y$ transformation so plug in for $v$. Doing this gives,

\[\begin{array}{c} \displaystyle - \frac{5}{2} \le y \le 0\\ \displaystyle - \frac{5}{2} \le \frac{1}{two}\left( {u - \left( {\frac{u}{2} + 2} \right)} \right) \le 0\\ \displaystyle - v \le \frac{u}{2} - 2 \le 0\\ \displaystyle - 3 \le \frac{u}{2} \le 2\\ - vi \le u \le 4\end{array}\]

So, again nosotros go the range of $u$'southward we look to get from the graph. Once we have those the advisable range of $v$ can be found from the equation itself as follows,

\[\brainstorm{array}{c} \displaystyle - 6 \le u \le 4\\ - three \le \frac{u}{2} \le 2\\ \displaystyle - 1 \le \frac{u}{ii} + 2 \le 4\\ - 1 \le five \le 4\end{assortment}\]

Basically, start with the range of $u$'south and "build up" the equation for the side and nosotros go the range of $v$'southward for this side.

Then, we now know how to go ranges of $u$ and/or $v$ for new equations under a transformation. However, this is not something that is washed terribly often just information technology is a useful skill to have in case it does arise somewhere.

At present that we've seen a couple of examples of transforming regions we demand to now talk most how we really do modify of variables in the integral. We will showtime with double integrals. In gild to change variables in a double integral we will need the Jacobian of the transformation. Hither is the definition of the Jacobian.

Definition

The Jacobian of the transformation $x = g\left( {u,v} \right)$, $y = h\left( {u,v} \correct)$ is

\[\frac{{\partial \left( {x,y} \correct)}}{{\fractional \left( {u,five} \right)}} = \left| {\brainstorm{assortment}{*{twenty}{c}}{ \displaystyle \frac{{\partial x}}{{\partial u}}}&{ \displaystyle \frac{{\partial x}}{{\partial v}}}\\{ \displaystyle \frac{{\partial y}}{{\partial u}}}& \displaystyle {\frac{{ \partial y}}{{\partial v}}}\terminate{array}} \right|\]

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

\[\left| {\begin{array}{*{20}{c}}a&b\\c&d\finish{array}} \right| = advertisement - bc\]

Therefore, some other formula for the determinant is,

\[\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \correct)}} = \left| {\begin{assortment}{*{20}{c}}{ \displaystyle \frac{{\partial x}}{{\partial u}}}&\displaystyle {\frac{{\partial ten}}{{\partial five}}}\\{ \displaystyle \frac{{\partial y}}{{\fractional u}}}&{ \displaystyle \frac{{\partial y}}{{\partial v}}}\end{array}} \right| = \frac{{\partial x}}{{\partial u}}\frac{{\partial y}}{{\partial 5}} - \displaystyle \frac{{\partial x}}{{\partial v}}\frac{{\partial y}}{{\partial u}}\]

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

Modify of Variables for a Double Integral

Suppose that we desire to integrate $f\left( {x,y} \right)$ over the region $R$. Under the transformation $x = g\left( {u,5} \correct)$, $y = h\left( {u,v} \correct)$ the region becomes $S$ and the integral becomes,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{South}{{f\left( {one thousand\left( {u,v} \correct),h\left( {u,v} \correct)} \right)\left| {\frac{{\partial \left( {ten,y} \correct)}}{{\partial \left( {u,v} \right)}}} \right|\,d\overline{A}}}\]

Note that we use $d\overline{A}$ in the $u$/$v$ integral in a higher place to denote that it volition exist in terms of $du$ and $dv$ once we convert to ii single integrals rather than the $dx$ and $dy$ we are used to using for $dA$. This is notational only and we by and large just employ $dA$ for both and just make sure to remember that the "new" $dA$ is in terms of $du$ and $dv$.

Besides note that we are taking the absolute value of the Jacobian.

If nosotros look just at the differentials in the to a higher place formula we can as well say that

\[dA = \left| {\frac{{\partial \left( {x,y} \correct)}}{{\partial \left( {u,v} \right)}}} \correct|\,d\overline{A}\]

Example three Show that when changing to polar coordinates we have $dA = r\,dr\,d\theta $

Show Solution

So, what nosotros are doing hither is justifying the formula that we used dorsum when we were integrating with respect to polar coordinates. All that we need to do is use the formula in a higher place for $dA$.

The transformation here is the standard conversion formulas,

\[10 = r\cos \theta \hspace{0.25in}\hspace{0.25in}y = r\sin \theta \]

The Jacobian for this transformation is,

\[\begin{align*}\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}} & = \left| {\begin{assortment}{*{xx}{c}}{\displaystyle \frac{{\fractional x}}{{\partial r}}}&{\displaystyle \frac{{\partial x}}{{\partial \theta }}}\\\displaystyle {\frac{{\partial y}}{{\partial r}}}&\displaystyle {\frac{{\partial y}}{{\partial \theta }}}\end{array}} \right|\\ & = \left| {\brainstorm{array}{*{20}{c}}{\cos \theta }&{ - r\sin \theta }\\{\sin \theta }&{r\cos \theta }\end{array}} \right|\\ & = r{\cos ^ii}\theta - \left( { - r{{\sin }^ii}\theta } \correct)\\ & = r\left( {{{\cos }^ii}\theta + {{\sin }^two}\theta } \right)\\ & = r\end{marshal*}\]

We then get,

\[dA = \left| {\frac{{\partial \left( {x,y} \correct)}}{{\partial \left( {r,\theta } \right)}}} \right|\,dr\,d\theta = \left| r \right|dr\,d\theta = r\,dr\,d\theta \]

And then, the formula we used in the department on polar integrals was correct.

Now, let'due south exercise a couple of integrals.

Example 4 Evaluate $\displaystyle \iint\limits_{R}{{x + y\,dA}}$ where $R$ is the trapezoidal region with vertices given by $\left( {0,0} \right)$, $\left( {5,0} \right)$, $\displaystyle \left( {\frac{v}{2},\frac{five}{2}} \right)$ and $\displaystyle \left( {\frac{5}{2}, - \frac{5}{2}} \right)$ using the transformation $x = 2u + 3v$ and $y = 2u - 3v$.

Show Solution

Offset, let'south sketch the region $R$ and determine equations for each of the sides.

$This is the 2D graph on a xy axis system of a diamond with vertices (0,0), $\left( \frac{5}{2}, \frac{5}{2} \right)$, (5,0) and $\left( \frac{5}{2}, -\frac{5}{2} \right)$. The top left edge of the triangle is given by y=x. The top right edge is given by y=-x+5. The bottom left edge of the triangle is given by y=-x. The bottom right edge is given by y=x-5. The diamond has been shaded in.$

Each of the equations was plant by using the fact that we know two points on each line (i.e. the 2 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.

Permit'southward apply the transformation and see what we get. Nosotros'll practise this past plugging the transformation into each of the equations above.

Allow'southward start the process off with $y = x$.

\[\begin{align*}2u - 3v & = 2u + 3v\\ 6v & = 0\\ v & = 0\end{align*}\]

Transforming $y = - 10$ is similar.

\[\begin{align*}2u - 3v & = - \left( {2u + 3v} \correct)\\ 4u & = 0\\ u & = 0\end{align*}\]

Next, we'll transform $y = - 10 + 5$.

\[\brainstorm{align*}2u - 3v & = - \left( {2u + 3v} \right) + v\\ 4u & = five\\ u & = \frac{5}{4}\end{marshal*}\]

Finally, allow's transform $y = 10 - 5$.

\[\brainstorm{align*}2u - 3v & = 2u + 3v - 5\\ - 6v & = - v\\ v & = \frac{5}{6}\finish{marshal*}\]

The region $Due south$ is then a rectangle whose sides are given by $u = 0$, $v = 0$, $u = \frac{5}{4}$ and $v = \frac{5}{vi}$ and so the ranges of $u$ and $v$ are,

\[0 \le u \le \frac{5}{four}\hspace{0.25in}\hspace{0.25in}0 \le v \le \frac{v}{6}\]

Next, we demand the Jacobian.

\[\frac{{\partial \left( {x,y} \correct)}}{{\partial \left( {u,v} \right)}} = \left| {\begin{array}{*{twenty}{c}}2&3\\2&{ - 3}\terminate{array}} \right| = - six - half-dozen = - 12\]

The integral is then,

\[\begin{align*}\iint\limits_{R}{{x + y\,dA}} & = \int_{{\,0}}^{{\,\frac{v}{6}}}{{\int_{{\,0}}^{{\,\frac{5}{4}}}{{\left( {\left( {2u + 3v} \right) + \left( {2u - 3v} \right)} \right)\left| { - 12} \correct|}}\,d\overline{A}}}\\ & = \int_{{\,0}}^{{\,\frac{5}{six}}}{{\int_{{\,0}}^{{\,\frac{5}{iv}}}{{48u}}\,d\overline{A}}}\\ & = \int_{{\,0}}^{{\,\frac{5}{6}}}{{\left. {24{u^2}} \correct|_0^{\frac{v}{iv}}\,dv}}\\ & = \int_{{\,0}}^{{\,\frac{five}{six}}}{{\frac{{75}}{2}\,dv}}\\ & = \left. {\frac{{75}}{two}v} \right|_0^{\frac{five}{half-dozen}}\\ & = \frac{{125}}{iv}\finish{align*}\]

Example v Evaluate $\displaystyle \iint\limits_{R}{{{x^2} - xy + {y^two}\,dA}}$ where $R$ is the ellipse given by ${x^2} - xy + {y^ii} \le 2$ and using the transformation $x = \sqrt 2 \,u - \sqrt {\frac{2}{3}} \,five$, $y = \sqrt 2 \,u + \sqrt {\frac{2}{3}} \,v$.

Prove Solution

Before nosotros proceed with this problem. Let's do a quick graph of the boundary of the region $R$. We claimed that it is an ellipse, but is clearly not in "standard" form. Here is the boundary of $R$.

So, it is an ellipse, just one that is at an bending rather than symmetric about the $x$ and $y$-centrality as we are used to dealing with.

Likewise, note that we used "$ \le 2$" when "defining" $R$ to brand it clear that we are using both the actual ellipse itself as well as the interior of the ellipse for $R$.

Okay, permit's keep with the problem.

The kickoff affair to practise is to plug the transformation into the equation for the ellipse to see what the region transforms into.

\[\begin{align*}2 & = {10^2} - xy + {y^2}\\ & = {\left( {\sqrt ii \,u - \sqrt {\frac{two}{iii}} \,v} \right)^2} - \left( {\sqrt 2 \,u - \sqrt {\frac{2}{3}} \,v} \right)\left( {\sqrt 2 \,u + \sqrt {\frac{two}{3}} \,v} \right) + {\left( {\sqrt 2 \,u + \sqrt {\frac{two}{3}} \,5} \right)^2}\\ & = 2{u^2} - \frac{four}{{\sqrt 3 }}uv + \frac{2}{3}{v^2} - \left( {2{u^two} - \frac{2}{3}{v^2}} \right) + 2{u^2} + \frac{four}{{\sqrt 3 }}uv + \frac{ii}{three}{v^2}\\ & = 2{u^2} + 2{v^2}\terminate{align*}\]

Or, upon dividing by two we see that the equation describing $R$ transforms into

\[{u^2} + {v^2} = ane\]

or the unit circle. Once more, this will be much easier to integrate over than the original region.

Note too that we've shown that the function that nosotros're integrating is

\[{x^2} - xy + {y^2} = two\left( {{u^2} + {five^2}} \right)\]

in terms of $u$ and $5$ and then we won't have to redo that work when the time to do the integral comes around.

Finally, we need to find the Jacobian.

\[\frac{{\partial \left( {x,y} \correct)}}{{\partial \left( {u,v} \right)}} = \left| {\begin{assortment}{*{20}{c}}{\sqrt 2 }&{ - \sqrt {\frac{2}{iii}} }\\{\sqrt 2 }&{\sqrt {\frac{2}{3}} }\end{array}} \right| = \frac{2}{{\sqrt three }} + \frac{2}{{\sqrt 3 }} = \frac{4}{{\sqrt 3 }}\]

The integral is then,

\[\iint\limits_{R}{{{ten^two} - xy + {y^2}\,dA}} = \iint\limits_{Southward}{{2\left( {{u^two} + {5^2}} \right)\left| {\frac{4}{{\sqrt iii }}} \right|\,du\,dv}}\]

Before proceeding a give-and-take of caution is in order. Exercise not brand the mistake of substituting ${10^2} - xy + {y^2} = 2$ or ${u^2} + {5^2} = one$ in for the integrand. These equations are just valid on the boundary of the region and nosotros are looking at all the points interior to the boundary also and for those points neither of these equations will exist truthful!

At this point we'll notation that this integral volition be much easier in terms of polar coordinates so to stop the integral out will convert to polar coordinates.

\[\begin{align*}\iint\limits_{R}{{{x^ii} - xy + {y^two}\,dA}} & = \iint\limits_{S}{{2\left( {{u^2} + {five^ii}} \right)\left| {\frac{4}{{\sqrt 3 }}} \correct|\,du\,dv}}\\ & = \frac{8}{{\sqrt 3 }}\int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,1}}{{\left( {{r^two}} \right)r\,dr}}\,d\theta }}\\ & = \frac{8}{{\sqrt 3 }}\int_{0}^{{2\pi }}{{\left. {\frac{1}{4}{r^4}} \right|_0^ane\,d\theta }}\\ & = \frac{8}{{\sqrt 3 }}\int_{0}^{{2\pi }}{{\frac{1}{4}\,d\theta }}\\ & = \frac{{4\pi }}{{\sqrt 3 }}\cease{align*}\]

Allow'due south at present briefly wait at triple integrals. In this case we will again commencement with a region $R$ and use the transformation $ten = yard\left( {u,v,w} \right)$, $y = h\left( {u,v,w} \right)$, and $z = 1000\left( {u,v,due west} \right)$ to transform the region into the new region $S$. To do the integral we will need a Jacobian, but equally we did with double integrals. Hither is the definition of the Jacobian for this kind of transformation.

\[\frac{{\partial \left( {x,y,z} \right)}}{{\partial \left( {u,v,w} \right)}} = \left| {\begin{array}{*{20}{c}}{ \displaystyle \frac{{\partial x}}{{\partial u}}}& \displaystyle {\frac{{\partial x}}{{\partial 5}}}& \displaystyle {\frac{{\partial x}}{{\fractional w}}}\\ \displaystyle {\frac{{\partial y}}{{\fractional u}}}& \displaystyle {\frac{{\fractional y}}{{\partial v}}}& \displaystyle {\frac{{\fractional y}}{{\fractional w}}}\\ \displaystyle{\frac{{\partial z}}{{\partial u}}}& \displaystyle {\frac{{\partial z}}{{\fractional v}}}& \displaystyle {\frac{{\partial z}}{{\partial w}}}\end{array}} \right|\]

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 cantankerous products back in Calculus Two. If you need a refresher on how to compute them you should go back and review that section.

The integral under this transformation is,

\[\iiint\limits_{R}{{f\left( {x,y,z} \right)\,dV}} = \iiint\limits_{Southward}{{f\left( {chiliad\left( {u,v,w} \right),h\left( {u,five,due west} \correct),k\left( {u,five,west} \correct)} \right)\left| {\frac{{\partial \left( {x,y,z} \right)}}{{\fractional \left( {u,five,due west} \correct)}}} \right|\,d\overline{V}}}\]

As with double integrals we used $d\overline{V}$ in the $u$/$v$/$w$ integral higher up to remind ourselves that we will demand to employ $du$, $dv$ and $dw$ when converting to single integrals. Once more, this is only notation and is ordinarily written as just $dV$.

We tin can look at simply the differentials and annotation that we must have

\[dV = \left| {\frac{{\partial \left( {x,y,z} \right)}}{{\partial \left( {u,v,w} \right)}}} \right|\,d\overline{Five}\]

We're not going to do any integrals hither, only let's verify the formula for $dV$ for spherical coordinates.

Example six Verify that $dV = {\rho ^two}\sin \varphi \,d\rho \,d\theta \,d\varphi $ when using spherical coordinates.

Show Solution

Here the transformation is merely the standard conversion formulas.

\[ten = \rho \sin \varphi \cos \theta \hspace{0.25in}y = \rho \sin \varphi \sin \theta \hspace{0.25in}z = \rho \cos \varphi \]

The Jacobian is,

\[\brainstorm{align*}\frac{{\partial \left( {10,y,z} \right)}}{{\partial \left( {\rho ,\theta ,\varphi } \right)}} & = \left| {\begin{array}{*{20}{c}}{\sin \varphi \cos \theta }&{ - \rho \sin \varphi \sin \theta }&{\rho \cos \varphi \cos \theta }\\{\sin \varphi \sin \theta }&{\rho \sin \varphi \cos \theta }&{\rho \cos \varphi \sin \theta }\\{\cos \varphi }&0&{ - \rho \sin \varphi }\finish{array}} \right|\\ & = - \rho ^ii\sin ^3\varphi \cos^2\theta - \rho^2 \sin \varphi \cos ^2\varphi \sin^2\theta + 0\\ & \hspace{0.75in} - \rho^two \sin^3 \varphi \sin^2 \theta - 0 - \rho^2 \sin \varphi \cos^two\varphi {\cos ^2}\theta \\ & = - {\rho ^2}\sin^iii\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - \rho^2 \sin \varphi \cos^2 \varphi \left( {{{\sin }^2}\theta + {{\cos }^two}\theta } \right)\\ & = - {\rho ^2}{\sin ^3}\varphi - \rho ^ii \sin \varphi \cos^two \varphi \\ & = - {\rho ^2}\sin \varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \correct)\\ & = - {\rho ^2}\sin \varphi \finish{align*}\]

Finally, $dV$ becomes,

\[dV = \left| { - {\rho ^2}\sin \varphi } \right|\,d\rho \,d\theta \,d\varphi = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi \]

Recall that we restricted $\varphi $ to the range $0 \le \varphi \le \pi $ for spherical coordinates so nosotros know that $\sin \varphi \ge 0$ and so we don't need the absolute value confined on the sine.

We will exit it to y'all to check the formula for $dV$ for cylindrical coordinates if y'all'd like to. It is a much easier formula to bank check.