calculuspowerup.com » Multiple Integration http://calculuspowerup.com Fri, 25 Jun 2010 05:10:05 +0000 http://wordpress.org/?v=2.9 en hourly 1 Change of Coordinates and the Jacobian http://calculuspowerup.com/change-of-coordinates-and-the-jacobian/ http://calculuspowerup.com/change-of-coordinates-and-the-jacobian/#comments Mon, 13 Apr 2009 00:42:10 +0000 Merrin http://calculuspowerup.com/?p=1221

Changing the variables to a new coordinate system sometimes can simplify an integration. Recall the coordinate transformation we are already familiar with, the u substitution. The new variable is u and the scale factor of the transformation is $latex \frac{du}{dx}$. Multiple integrals also involve a scale factor called the Jacobian which corrects for the relative sizes of area or volume elements between coordinate systems. Only a few of the major coordinate systems: Cartesian coordinates, polar coordinates, cylindrical coordinates, and spherical coordinates are referred to in this book.

The general formula for a two dimensional Jacobian between Cartesian coordinates x, y and generalized coordinates u, v is

J(u,v)=\cfrac{\partial (u,v)}{\partial (x,y)} =\left|\begin{array}{cc} {\cfrac{\partial x}{\partial u} } & {\cfrac{\partial x}{\partial v} } \\ {\cfrac{\partial y}{\partial u} } & {\cfrac{\partial y}{\partial v} } \end{array}\right| \iint \nolimits _{R}f(x,y)dxdy =\iint \nolimits _{R}f(x(u,v),y(u,v))\left|J(u,v)\right|dudv

The general formula for a three dimensional Jacobian between Cartesian coordinates x, y, z and generalized coordinates u, v, w is given as

J(u,v,w)=\cfrac{\partial (u,v,w)}{\partial (x,y,z)} =\left|\begin{array}{ccc} {\cfrac{\partial x}{\partial u} } & {\cfrac{\partial x}{\partial v} } & {\cfrac{\partial x}{\partial w} } \\ {\cfrac{\partial y}{\partial u} } & {\cfrac{\partial y}{\partial v} } & {\cfrac{\partial y}{\partial w} } \\ {\cfrac{\partial z}{\partial u} } & {\cfrac{\partial z}{\partial v} } & {\cfrac{\partial z}{\partial w} } \end{array}\right|
\iiint \nolimits _{V}f(x,y,z)dxdy =\iiint \nolimits _{V}f(x(u,v,w),y(u,v,w),z(u,v,w))\left|J(u,v,w)\right|dudvdw

Example 1. Define polar coordinates and calculate the Jacobian for the transformation from Cartesian coordinates to polar coordinates.

Solution 1. Polar coordinates are defined as

x=r\cos \theta \quad \, y=r\sin \theta

A polar coordinate lies in a plane specified by a radius and an angle.

polar

We need to find the factor Jacobian factor $latex \left|J(r,\theta )\right|$ which allows us to convert an integral from (x,y) coordinates to $latex (r,\theta)$ coordinates .

\iint \nolimits _{R}f(x,y)dxdy =\iint \nolimits _{R}f(r\cos \theta ,r\sin \theta )\left|J(r,\theta )\right|drd\theta

The formula for the Jacobian for polar coordinates is

{J(r,\theta )}= {\left|\begin{array}{cc} {\cfrac{\partial x}{\partial r} } & {\cfrac{\partial x}{\partial \theta } } \\ {\cfrac{\partial y}{\partial r} } {\cfrac{\partial y}{\partial \theta } } \end{array}\right|=\left|\begin{array}{cc} {\cos \theta } {-r\sin \theta } \\ {\sin \theta } {r\cos \theta } \end{array}\right|} \\ = {r((\cos \theta )^{2} +(\sin \theta )^{2} )} \\ = {r}

\iint \nolimits _{R}f(x,y)dxdy =\iint \nolimits _{R}f(r\cos \theta ,r\sin \theta )rdrd\theta

Integrands with polar symmetry are generally easier to integrate in polar coordinates. The most general polar region that can be integrated in one piece is

\displaystyle \int _{\theta _{1} }^{\theta _{2} }\displaystyle \int _{r{\kern 1pt} _{1} (\theta )}^{r_{2} (\theta )}f(r,\theta )rdrd\theta

The integration region lies between the two curves

R:\theta _{1} \le \theta \le \theta _{2} , r_{1} (\theta )\le r\le r_{2} (\theta )

This factor of r as the Jacobian is not at all strange. After all, the dimension of dxdy is length squared. The dimension of $latex drd\theta $ is only length. We would expect another factor of length to be necessary for the two integrals to be equal dimensionally. This is what the Jacobian exactly provides.

Example 2. Calculate the area of a circle in Cartesian coordinates and in polar coordinates using the Jacobian

Solution 2. To set up an integral in two dimensions that is the area of a circle we write.

{A=\displaystyle \int _{x=-R}^{x=R}\displaystyle \int _{y=-\sqrt{R^{2} -x^{2} } }^{y=\sqrt{R^{2} -x^{2} } }dxdy =4\displaystyle \int _{0}^{R}\sqrt{R^{2} -x^{2} } dx } \\ {x=R\sin \theta \quad dx=R\cos \theta } \\ {4\displaystyle \int _{0}^{\pi /2}\sqrt{R^{2} -R^{2} (\sin \theta )} R\cos \theta dx =4R^{2} \displaystyle \int _{0}^{\pi /2}(\cos \theta )^{2} d\theta } \\ {A=4\cfrac{1}{2} \cfrac{\pi }{2} R^{2} =\pi R^{2} }

We do the same integral but transform to polar coordinates. There is the function one in the integrand so we can simply write

A=\displaystyle \int _{0}^{2\pi }\displaystyle \int _{0}^{R}rdrd\theta = \left(\displaystyle \int _{0}^{2\pi }d\theta \right)\left(\displaystyle \int _{0}^{R}rdr \right)

This integral is said to be separable because there are no conflicts between variables being integrated in the integrand.

A=\displaystyle \int _{0}^{2\pi }\displaystyle \int _{0}^{R}rdrd\theta =(2\pi )(\cfrac{1}{2} R^{2} )=\pi R^{2}

The integration in polar coordinates is much easier, no trigonometric substitutions
required.

The area element of polar coordinates can be remembered as follows. Polar integration is like adding up annuli of width dr and a perimeter of $latex 2\pi r$. Convert $latex 2\pi $ into slices of $latex d\theta $ and we get $latex rdrd\theta $ as the area of an infinitesimal polar rectangle.

Example 3. Evaluate the following integral using polar coordinates.

I=\displaystyle \int _{-\infty }^{\infty }e^{-x^{2} } dx

Solution 3. We can solve the following integral by first squaring it and then making a transformation to polar coordinates.

{I^{2} } = {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2} } dx \displaystyle \int _{-\infty }^{\infty }e^{-y^{2} } dy=\displaystyle \int _{-\infty }^{\infty }\displaystyle \int _{-\infty }^{\infty }e^{-x^{2} -y^{2} } dxdy }

I^2=\displaystyle \int_0^{2\pi}\displaystyle \int_0^{\infty}e^{-r^2}rdrd\theta = 2\pi\cfrac{1}{2}\displaystyle \int_0^{\infty}e^{-u}du=\pi

So we can calculate the original integral by taking the square root and we have the result.

I=\displaystyle \int _{-\infty }^{\infty }e^{-x^{2} } dx =\sqrt{\pi }

The function integrated in the previous problem is called a Gaussian function and is also known as a Bell curve. The Gaussian function is important in statistics and physics. We have calculated the total area beneath this function. Although there is no simple indefinite integral for the Gaussian there is a well defined value for the definite integral which extends in both directions to infinity.

Example 4. Define cylindrical coordinates and then calculate the Jacobian for the transformation to cylindrical coordinates.

Solution 4. Cylindrical coordinates are defined relative to Cartesian coordinates as

x=r\cos \theta \quad y=r\sin \theta \quad z=z

Cylindrical coordinates are just polar coordinates plus a z direction.

cylindrical

To find the Jacobian for polar coordinates we will need a bigger matrix because there are three coordinates to consider.

J(r,\theta ,z)=\left|\begin{array}{ccc} {\cfrac{\partial x}{\partial r} } & {\cfrac{\partial x}{\partial \theta } } & {\cfrac{\partial x}{\partial z} } \\ {\cfrac{\partial y}{\partial r} } & {\cfrac{\partial y}{\partial \theta } } & {\cfrac{\partial y}{\partial z} } \\ {\cfrac{\partial z}{\partial r} } & {\cfrac{\partial z}{\partial \theta } } & {\cfrac{\partial z}{\partial z} } \end{array}\right|=\left|\begin{array}{ccc} {\cos \theta } & {-r\sin \theta } & {0} \\ {\sin \theta } & {\cos \theta } & {0} \\ {0} & {0} & {1} \end{array}\right|=r

It is not surprising that the Jacobian for cylindrical coordinates is the same as for polar coordinates since dz transforms into dz.

Example 5. Find the volume of a cylinder with radius R and height H.

Solution 5. We use cylindrical coordinates and find the volume to be

V=\displaystyle \int _{0}^{2\pi }d\theta \displaystyle \int _{0}^{R}rdr \displaystyle \int _{0}^{H}dz =(2\pi )(R^{2} /2)H=\pi R^{2} H

Example 6. Define spherical coordinates, and then calculate the Jacobian for spherical coordinates.

Solution 6. Spherical coordinates are defined relative to Cartesian coordinates as

x=r\sin \theta \cos \phi \quad y=r\sin \theta \sin \phi \quad z=r\cos \theta

There are two angles in spherical coordinates and a radial coordinate. The angular coordinate relative to the z axis is called the polar angle, like the north south pole. The angular coordinate in the xy plane is called the azimuthal angle. The radial coordinate lies on the ray emanating from the origin for the particular angle chosen.

spherical

We set up our calculation for the Jacobian as we have done before but this time using spherical coordinates.

J(r,\theta ,\phi )=\left|\begin{array}{ccc} {\cfrac{\partial x}{\partial r} } & {\cfrac{\partial x}{\partial \theta } } & {\cfrac{\partial x}{\partial \phi } } \\ {\cfrac{\partial y}{\partial r} } & {\cfrac{\partial y}{\partial \theta } } & {\cfrac{\partial y}{\partial \phi } } \\ {\cfrac{\partial z}{\partial r} } & {\cfrac{\partial z}{\partial \theta } } & {\cfrac{\partial z}{\partial \phi } } \end{array}\right|

Verify that the determinant gives the indicated value.

J(r,\theta ,\phi )=\left|\begin{array}{ccc} {\sin \theta \cos \phi } & {r\cos \theta \cos \phi } & {-r\sin \theta \sin \phi } \\ {\sin \theta \sin \phi } & {r\cos \theta \sin \phi } & {r\sin \theta \cos \phi } \\ {\cos \theta } & {-r\sin \theta } & {0} \end{array}\right|=r^{2} \sin \theta

Example 7. Calculate the volume of a sphere by first writing an integral in Cartesian coordinates or first in spherical coordinates.

Solution 7. We can set up an integral in spherical coordinates.

{\displaystyle \int _{0}^{2\pi }\displaystyle \int _{0}^{\pi }\displaystyle \int _{r=0}^{R}r^{2} \sin \theta d\theta d\phi } = {\displaystyle \int _{0}^{2\pi }d\phi \displaystyle \int _{0}^{\pi }\sin \theta d\theta \displaystyle \int _{0}^{R}r^{2} dr }

The Cartesian integral is a bit more complicated to evaluate, because the integrals are not separable.

{\displaystyle \int _{-R}^{R}\displaystyle \int _{-\sqrt{R^{2} -x^{2} } }^{\sqrt{R^{2} -x^{2} } }\displaystyle \int _{-\sqrt{R^{2} -x^{2} -y^{2} } }^{\sqrt{R^{2} -x^{2} -y^{2} } }dzdydx } = {8\displaystyle \int _{0}^{R}\displaystyle \int _{0 } }^{\sqrt{R^{2} -x^{2} } }\displaystyle \int _{0}^{\sqrt{R^{2} -x^{2} -y^{2} } }dzdydx

I have exhausted all possible symmetry in the integrals which are all even with the factor of 8 and halving the domains of each direction. This integration is starting to get painful. The next step is a trigonometric substitution of $latex y=\sqrt{R^{2} -x^{2} } \sin \theta $. I don’t feel like doing it so let us just convert the rest to polar coordinates and get the work done in a jiffy. Note that the conversion to polar coordinates makes the rest of the integral separable.

8\displaystyle \int _{0}^{\pi /2}\displaystyle \int _{0}^{R}\sqrt{R^{2} -r^{2} } rdrd\theta =-8(\cfrac{\pi }{2} )\cfrac{2}{3} \left. \cfrac{(R^{2} -r^{2} )^{3/2} }{2} \right|_{0}^{R} =\cfrac{4\pi }{3} R^{3}

It appears we have written down the correct Cartesian coordinates integral for the volume of a sphere. I leave it to you to fill in the details of the calculation in xyz land if you dare.

For this section, try to remember our Jacobian for cylindrical and polar coordinates is just r, and that our Jacobian for spherical coordinates is $latex r^{2} \sin \theta $. As an exercise, try to convert from the new coordinates systems back to Cartesian coordinates. Often integrals in spherical coordinates are separable over the two angles which is sometimes called the solid angle. This can be written as

{d\Omega =\sin \theta d\theta d\phi } \\ {\iint \nolimits _{}d \Omega =4\pi }

Solid angle is what one sees when looking around inside a spherical shell from the center. For a circle in 2D, we have 360 degrees view. In 3D, we have a view of $latex 4\pi $. A subsection of the total solid angle is just a limited integral which we now know how to find.

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