Home page for accesible maths 4 Double integrals: general regions and change of variable 4.1 Review of double integrals 4.3 Double integrals using polar coordinates

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4.2 Double integrals over triangular and more general regions

Specifying regions by vertical strips.

Let $y_{1}(x)$ and $y_{2}(x)$ be functions of $x$ with $y_{1}(x)\leq y_{2}(x)$ for $a\leq x\leq b$ . Then

R=\{(x,y):y_{1}(x)\leq y\leq y_{2}(x);a\leq x\leq b\}

defines the region that is bounded by the lines $x=a$ and $x=b$ and the graphs $y_{1}(x)$ and $y_{2}(x).$ It is natural to think of this region as being made up of narrow vertical strips from $y_{1}(x)$ up to $y_{2}(x)$ , which move from left to right covering the region as $x$ increases from $a$ to $b$ .

We can write

\int\!\!\!\int_{R}f(x,y)\,dxdy=\int_{a}^{b}\int_{y_{1}(x)}^{y_{2}(x)}f(x,y)dydx.

Specifying regions by horizontal strips.

Let’s now think of another region $S$ , which we describe differently. Let $x_{1}(x)$ and $x_{2}(x)$ be functions of $y$ with $x_{1}(y)\leq x_{2}(y)$ for $c\leq y\leq d$ . Then

S=\{(x,y):x_{1}(y)\leq x\leq x_{2}(y);c\leq y\leq d\}

defines the region that is bounded by the lines $y=c$ and $y=d$ and the graphs $x_{1}(y)$ and $x_{2}(y).$ It is natural to think of $S$ as being made up of narrow horizontal strips from $x_{1}(y)$ up to $x_{2}(y)$ , which move from bottom to top covering the region as $y$ increases from $c$ to $d$ .

We can write

\int\!\!\!\int_{S}f(x,y)\,dxdy=\int_{c}^{d}\int_{x_{1}(y)}^{x_{2}(y)}f(x,y)dxdy.

We can cover some regions by either horizontal or vertical strips.

Example. Let $T$ be the triangle that is bounded by the lines $x=0$ , $y=a$ and $y=x$ . Find the double integral $\int\!\!\!\int_{T}{{xdxdy}\over{\sqrt{x^{2}+y^{2}}}}.$

Note first that the given double integral is equal to

\int_{0}^{a}\!\!\!\int_{0}^{y}{{x\,dxdy}\over{\sqrt{x^{2}+y^{2}}}}\quad{\hbox{% and}}\quad\int_{0}^{a}\!\!\!\int_{x}^{a}{{x\,dydx}\over{\sqrt{x^{2}+y^{2}}}}.

For the double integral on the left-hand side, the inner integral is

$\displaystyle{\int_{0}^{y}\frac{x\,dx}{\sqrt{x^{2}+y^{2}}}=}$

Now the outer integral is $\int_{0}^{a}(\sqrt{2}-1)y\,dy=$

On the right-hand side, the inner integral is $\int_{x}^{a}\frac{x\,dy}{\sqrt{x^{2}+y^{2}}}$ , which we solve by the substitution $y=x\sinh u$ , so that

$\displaystyle{x^{2}+y^{2}=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\mbox{and}% \;\;\frac{dy}{du}=}$

Then the inner integral turns out to be

$\displaystyle{\int_{\sinh^{-1}1}^{\sinh^{-1}\frac{a}{x}}\frac{x^{2}\cosh u}{x% \cosh u}\,du=}$

which leaves us with a difficult (though not impossible) outer integral.

So clearly the integral on the left-hand side was the easier choice.

Integrals on region bounded by parabola.

Example. Find $\int\!\!\!\int_{R}xy\,dxdy$ , where $R$ is the region that is bounded by $x=0$ , $y=2$ and the graph of $y=\sqrt{x}.$

The curve $y=\sqrt{x}$ meets the $y$ -axis at $(0,0)$ , and meets the line $y=2$ at $(4,2)$ . Thus the region $R$ is given by $0\leq x\leq 4$ , $\sqrt{x}\leq y\leq 2$ . Therefore the integral is

$\displaystyle{\int_{0}^{4}\!\int_{\sqrt{x}}^{2}\,xy\,dydx=}$

Area of a disc.

Recall that for a bounded region $R$ in the plane, the area of $R$ is given by $\int\!\!\!\int_{R}dxdy.$ Often this reduces to the integral of a single variable.

Example. Find the area of the disc $D$ $\{(x,y):x^{2}+y^{2}\leq r^{2}\}$ .

The disc itself corresponds to the points $(x,y)$ that satisfy

-\sqrt{r^{2}-y^{2}}\leq x\leq\sqrt{r^{2}-y^{2}}\quad{\hbox{and}}\quad-r\leq y% \leq r.

Hence we can express the area as $\int\!\!\!\int_{D}dxdy=2\int_{-r}^{r}\sqrt{r^{2}-y^{2}}dy.$ To evaluate this, we let $y=r\sin\theta$ and obtain the expected value of $\pi r^{2}$ after a little reduction.

2\int_{-r}^{r}\sqrt{r^{2}-y^{2}}\,dy=2r^{2}\int_{-\pi/2}^{\pi/2}\cos^{2}\theta% \,d\theta=r^{2}\int_{-\pi/2}^{\pi/2}\bigl{(}\cos 2\theta+1\bigr{)}\,d\theta

This equals

as expected.

In the next section, we will see how to compute areas of circular regions very easily by changing variables (from Cartesian to polar coordinates) in double integrals.