Home page for accesible maths 4 Double integrals: general regions and change of variable 4.3 Double integrals using polar coordinates 5 Volumes and surface areas of solids

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4.4 Change of variable for double integrals: the general case

Let $u$ , $v$ be new coordinates, and $x=x(u,v),\quad y=y(u,v)$ . The problem is to calculate a double integral (i.e. a volume) in terms of $u$ and $v$ . For this, we need an estimate of the area of the small region bounded by the curves corresponding to values $u,\;u+\delta u,\;v,\;v+\delta v$ .

In the diagram,

P

is the point

[x(u,v),y(u,v)]

and

A

is the point

[x(u+\delta u,v),\;y(u+\delta u,v)]

. Recall that

x(u+\delta u,v)-x(u,v)\approx x_{u}\;\delta u,

in which

x_{u}

is evaluated at

(u,v)

. Similarly for

y

. Hence the vector

\overrightarrow{PA}

is approximately

(x_{u}\;\delta u,\;y_{u}\;\delta u)

. Similarly

\overrightarrow{PB}\approx(x_{v}\;\delta v,\;y_{v}\;\delta v)

Lemma 4.6

Let S be a parallelogram with sides given by the vectors $(x_{1},y_{1})$ and $(x_{2},y_{2})$ . Then the area of S is $|x_{1}y_{2}-x_{2}y_{1}|$ .

Proof

Recall from Chapter 1 that the area is given by $|(x_{1},y_{1},0)\times(x_{2},y_{2},0)|$ , which equals $|(0,0,x_{1}y_{2}-x_{2}y_{1})|=|x_{1}y_{2}-x_{2}y_{1}|$ . $\square$

So the area $\delta A$ between the curves $u,\;u+\delta u,\;v,\;v+\delta v$ is approximately

which is equal to $|x_{u}y_{v}-x_{v}y_{u}|\;\delta u\;\delta v$ . Its contribution to the volume is $\delta A\times f[x(u,v),y(u,v)]$ . The expression

x_{u}y_{v}-x_{v}y_{u}=\left|\begin{array}[]{ll}x_{u}&x_{v}\\ y_{u}&y_{v}\end{array}\right|

is called the Jacobian $\displaystyle{\frac{\partial(x,y)}{\partial(u,v)}}$ .

Conclusion. We have

\int\!\int_{R}f(x,y)\;dx\;dy=\int\!\int_{R^{\prime}}f[x(u,v),y(u,v)]\left|% \frac{\partial(x,y)}{\partial(u,v)}\right|\;du\;dv,

where $R^{\prime}$ is the same region expressed in terms of u and v.

Polar coordinates. Since $x=r\cos\theta$ and $y=r\sin\theta$ , we have

\frac{\partial(x,y)}{\partial(r,\theta)}=\left|\begin{array}[]{rr}\cos\theta&-% r\sin\theta\\ \sin\theta&r\cos\theta\end{array}\right|=r(\cos^{2}\theta+\sin^{2}\theta)=r,

so the new result agrees with our earlier one for this case.

Sometimes it is more convenient to define $u$ and $v$ in terms of $x$ and $y$ (this is only valid if, in principle, we can solve to express $x$ and $y$ uniquely in terms of $u$ and $v$ ). As we saw in section 3,

\left(\begin{array}[]{cc}\displaystyle{\frac{\partial u}{\partial x}}&% \displaystyle{\frac{\partial u}{\partial y}}\\ \displaystyle{\frac{\partial v}{\partial x}}&\displaystyle{\frac{\partial v}{% \partial y}}\end{array}\right)\left(\begin{array}[]{cc}\displaystyle{\frac{% \partial x}{\partial u}}&\displaystyle{\frac{\partial x}{\partial v}}\\ \displaystyle{\frac{\partial y}{\partial u}}&\displaystyle{\frac{\partial y}{% \partial v}}\end{array}\right)=\left(\begin{array}[]{cc}1&0\\ 0&1\end{array}\right).

It follows that

\frac{\partial(u,v)}{\partial(x,y)}\>\frac{\partial(x,y)}{\partial(u,v)}=1,

so that the required Jacobian $\displaystyle{\frac{\partial(x,y)}{\partial(u,v)}}$ is simply the reciprocal of $\displaystyle{\frac{\partial(u,v)}{\partial(x,y)}}$ .

Example 4.7

Evaluate $\int\!\int_{R}(x^{2}-y^{2})\>dx\>dy$ , where $R$ is the rectangle bounded by $x+y=0$ , $x+y=2$ , $y=x$ and $y=x-1$ .

$\quad$

Example 4.8

Let $R$ be the region bounded by $xy=a$ , $xy=b$ , $y^{2}=px$ and $y^{2}=qx$ , where $0<a<b$ and $0<p<q$ . Find the area of $R$ , and evaluate $\int\!\int_{R}y^{3}\>dx\>dy$ .