5.2 Double integrals over general regions  (Page 2/12)

Describing a region as type i and also as type ii

Consider the region in the first quadrant between the functions $y=\sqrt{x}$ and $y=x^3$ ( [link] ). Describe the region first as Type I and then as Type II.

When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region $D$ is bounded above by $y=\sqrt{x}$ and below by $y=x^3$ in the interval for $x\text{in}\left[0,1\right].$ Hence, as Type I, $D$ is described as the set $\left\{\left(x,y\right)|0\le x\le 1,x^3\le y\le \sqrt{x}\right\}.$

However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region $D$ is bounded on the left by $x=y^2$ and on the right by $x=\sqrt[3]{y}$ in the interval for y in $[0,1].$ Hence, as Type II, $D$ is described as the set $\left\{\left(x,y\right)|0\le y\le 1,y^2\le x\le \sqrt[3]{y}\right\}.$

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Double integrals over nonrectangular regions

To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.

The right-hand side of this equation is what we have seen before, so this theorem is reasonable because $R$ is a rectangle and ${\displaystyle \underset{R}{\iint }g\left(x,y\right)}dA$ has been discussed in the preceding section. Also, the equality works because the values of $g\left(x,y\right)$ are $0$ for any point $\left(x,y\right)$ that lies outside $D,$ and hence these points do not add anything to the integral. However, it is important that the rectangle $R$ contains the region $D.$

As a matter of fact, if the region $D$ is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle $R$ containing the region.

The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate $f\left(x,y\right)$ with $x$ being held constant and the limits of integration being $g_1\left(x\right)\text{and}{g}_2\left(x\right).$ In the inner integral in the second expression, we integrate $f\left(x,y\right)$ with $y$ being held constant and the limits of integration are $h_1\left(x\right)\text{and}{h}_2\left(x\right).$