Change of Variable -- the Jacobian

Another technique that can sometimes be useful when trying to evaluate a double (or triple etc) integral generalise the familiar method of integration by substitution.

Assume we have a change of variable x = x(u, v) and y = y(u, v). Suppose that the region S' in the uv - plane is transformed to a region S in the xy - plane under this transformation. Define the Jacobian of the transformation as

J(u, v) = $$\begin{vmatrix} \displaystyle \frac{\partial x}{\partial u}&\dis... ...tial y}{\partial u} &\displaystyle \frac{\partial y}{\partial v} \end{vmatrix}$$ = $${\frac{{\partial(x,y)}}{{\partial(u,v)}}}$$.

It turns out that this correctly describes the relationship between the element of area dx dy and the corresponding area element du dv.

With this definition, the change of variable formula becomes:

$$\int$$$$\int_{S}^{}$$f (x, y) dx dy = $$\int$$$$\int_{S}{^\prime}$$f (x(u, v), y(u, v))$$\left\vert\vphantom{J(u,v)}\right.$$J(u, v)$$\left.\vphantom{J(u,v)}\right\vert$$ du dv.

Note that the formula involves the modulus of the Jacobian.

Example 9.9   Find the area of a circle of radius R.

Solution. Let A be the disc centred at 0 and radius R. The area of A is thus $$\int$$$$\int_{A}^{}$$dx dy. We evaluate the integral by changing to polar coordinates, so consider the usual transformation x = r cos$\theta$, y = r sin$\theta$ between Cartesian and polar co-ordinates. We first compute the Jacobian;

$${\frac{{\partial x}}{{\partial r}}}$$ = cos$$\theta$$,        $${\frac{{\partial y}}{{\partial r}}}$$ = sin$$\theta$$,        $${\frac{{\partial x}}{{\partial \theta}}}$$ = - r sin$$\theta$$,        $${\frac{{\partial y}}{{\partial \theta}}}$$ = r cos$$\theta$$.

Thus

J(r,$$\theta$$) = $$\begin{vmatrix} \displaystyle \displaystyle \frac{\partial x}{\p... ...y}{\partial r} &\displaystyle \frac{\partial y}{\partial \theta} \end{vmatrix}$$ = $$\begin{vmatrix} \cos\theta&-r\sin\theta\\ \sin \theta &r \cos \theta \end{vmatrix}$$ = r(cos²$$\theta$$ + sin²$$\theta$$) = r.

We often write this result as

dA = dx dy = r dr d$$\theta$$

Using the change of variable formula, we have

$$\int$$$$\int_{A}^{}$$dx dy = $$\int$$$$\int_{A}^{}$$| J(r,$$\theta$$)| dr d$$\theta$$ = $$\int_{0}^{R}$$$$\int_{0}^{{2\pi}}$$r dr d$$\theta$$ = 2$$\pi$$$${\frac{{R^2}}{{2}}}$$.

We thus recover the usual area of a circle.

Note that the Jacobian J(r,$\theta$) = r > 0, so we did indeed take the modulus of the Jacobian above.

Example 9.10   Find the volume of a ball of radius 1.

Solution. Let V be the required volume. The ball is the set {(x, y, z) | x² + y² + z²$\le$1}. It can be thought of as twice the volume enclosed by a hemisphere of radius 1 in the upper half plane, and so

V = 2$$\int_{D}^{}$$$$\sqrt{{1 - x^2 - y^2}}$$  dx dy

where the region of integration D consists of the unit disc {(x, y) | x² + y²$\le$1}. Although we can try to do this integration directly, the natural co-ordinates to use are plane polars, and so we instead do a change of variable first. As in 9.9, if we write x = r cos$\theta$, y = r sin$\theta$, we have dx dy = r dr d$\theta$. Thus

$$\int_{D}^{} \sqrt{{1 - x^2 - y^2}} \int_{D}^{} \sqrt{{1 - r^2}} \theta$$ =

$$\int_{0}^{{2\pi}} \theta \int_{0}^{1} \sqrt{{1 - r^2}}$$ =

$$\pi \left[\vphantom{-\frac{(1-r^2)^{3/2}}{3}}\right. {\frac{{(1-r^2)^{3/2}}}{{3}}} \left.\vphantom{-\frac{(1-r^2)^{3/2}}{3}}\right]_{0}^{1}$$ =

$${\frac{{4\pi}}{{3}}}$$ =

Note that after the change of variables, the integrand is a product, so we are able to do the dr and d$\theta$ parts of the integral at the same time.

And finally, we show that the same ideas work in 3 dimensions. There are (at least) two co-ordinate systems in $\mathbb {R}$³ which are useful when cylindrical or spherical symmetry arises. One of these, cylindrical polars is given by the transformation

x = r cos$$\theta$$,        y = r sin$$\theta$$,        z = z,

and the Jacobian is easily calculated as

$${\frac{{\partial(x,y,z)}}{{\partial(r,\theta,z)}}}$$ = r    so    dV = dx dy dz = r dr d$$\theta$$ dz.

The second useful co-ordinate system is spherical polars with transformation

x = r sin$$\phi$$cos$$\theta$$,        y = r sin$$\phi$$sin$$\theta$$,        z = r cos$$\phi$$.

The transformation is illustrated in Fig 9.3.

Figure 9.3: The transformation from Cartesian to spherical polar co-ordinates.

It is easy to check that Jacobian of this transformation is given by

dV = r²sin$$\phi$$ dr d$$\phi$$ d$$\theta$$ = dx dy dz.

Example 9.11   The moment of inertia of a solid occupying the region R, when rotated about the z - axis is given by the formula

I = $$\int$$$$\int$$$$\int_{R}^{}$$(x² + y²)$$\rho$$ dV.

Calculate the moment of inertia about the z-axis of the solid of unit density which lies outside the cylinder of radius a, inside the sphere of radius 2a, and above the x - y plane.

Solution. Let I be the moment of inertia of the given solid about the z-axis. A diagram of a cross section of the solid is shown in Fig 9.4.

Figure 9.4: Cross section of the right hand half of the solid outside a cylinder of radius a and inside the sphere of radius 2a

We use cylindrical polar co-ordinates (r,$\theta$, z); the Jacobian gives dx dy dz = r dr d$\theta$ dz, so

$$\int_{0}^{{2\pi}} \theta \int_{a}^{{2a}} \int_{0}^{{\sqrt{4a^2 - r^2}}}$$ I =

$$\pi \int_{a}^{{2a}} \sqrt{{4a^2 - r^2}}$$ =

We thus have a single integral. Using the substitution u = 4a² - r², you can check that the integral evaluates to 22$\sqrt{{3}}$$\pi$a⁵/5.

Exercise 9.12   Show that

$$\int$$ $$\int$$ $$\int$$z² dxdydz = $${\frac{{4}}{{15}}}$$$$\pi$$

(where the integral is over the unit ball x² + y² + z² $\leq$ 1) first by using spherical polars, and then by doing the z integration first and using plane polars.