Chapter 26
By the end of this section, you should be able to answer the following questions:
It is often possible to represent a type I region as a union of type II regions, or a type II region as a union of type I regions. Why would we want to do that? In some cases, it may only be possible to integrate a function one way but not the other. In this section, we investigate this idea more closely.
under the paraboloid $z=x^2+y^2$ above the region $D$, where $D$ is bounded by $y=x^2$ and $y=2x$. Do the problem twice, first by taking $D$ to be a type I region, then by taking $D$ to be a type II.
Type I region with $x^2 \leq y \leq 2x.$
We need to find $a$ and $b$ by solving \[ \left\{\begin{array}{ll} y=x^2\\ y=2x \end{array}\right. \] for $x$.
Here we have $x^2=2x$ $\Ra x = 0, x = 2$ $\Ra a = 0, b = 2.$
Then $D= \left\{ (x,y) ~|~ 0\leq x\leq 2, ~x^2\leq y\leq 2x\right\}$.
| $I$ | $= \displaystyle \int_{x=0}^{x=2} \left[ \int_{y=x^2}^{y=2x} \left(x^2+y^2\right)dy \right]dx $ |
| $ = \displaystyle \int_{x=0}^{x=2} \left[ \left( x^2 y + \frac{y^3}{3} \right) \Bigg|_{y=x^2}^{y=2x} \right] dx $ | |
| $= \displaystyle \int_{x=0}^{x=2} \left[ x^2 (2x) + \frac{(2x)^3}{3} -\left( x^2 \left(x^2\right)+ \frac{\left(x^2\right)^3}{3} \right) \right]dx$ | |
| $= \displaystyle \int_{x=0}^{x=2} \left( -\frac{x^6}{3} - x^4 + \frac{14 x^3}{3} \right) dx $ | |
| $= \displaystyle \left( \frac{7 x^4}{6} - \frac{x^5}{5} - \frac{x^7}{21} \right)\Bigg|_{x=0}^{x=2} $ $= \displaystyle \frac{216}{35}.$ |
Type II region. We know that $y=2x$ and $y=x^2$. So $x= \frac{y}{2}$ and $x=\sqrt{y}$. Then $\frac{y}{2} \leq x\leq \sqrt{y}$.
We need to find $c$ and $d$ by solving \[ \left\{\begin{array}{ll} x=\sqrt{y}\\ x=\dfrac{y}{2} \end{array}\right. \] for $y$.
Here we have $\sqrt{y}=\frac{y}{2}$ $\Ra y = 0, y = 4$ $\Ra c = 0, d = 4.$
Then $D= \left\{ (x,y) ~|~ 0\leq y\leq 4, ~ \frac{y}{2} \leq x\leq \sqrt{y} \right\}$.
| $I$ | $= \displaystyle \int_\limits{y=0}^{y=4} \left[ \int_{x=\frac{y}{2}}^{x=\sqrt{y}} \left(x^2+y^2\right) dx \right]dy$ $ = \displaystyle \int_\limits{y=0}^{y=4} \left[ \left( \frac{x^3}{3}+y^2x \right) \Bigg|_{x=\frac{y}{2}}^{x=\sqrt{y}} \right] dy $ |
| $ = \displaystyle \int_\limits{y=0}^{y=4} \left[ \frac{\left(\sqrt{y}\right)^3}{3}+y^2\left(\sqrt{y}\right) - \left( \frac{\left(\frac{y}{2}\right)^3}{3}+y^2\left(\frac{y}{2}\right) \right) \right]dy $ | |
| $= \displaystyle \int_\limits{y=0}^{y=4} \left( y^{5/2} + \frac{y^{3/2}}{3} - 13 \frac{y^3}{24} \right) dy $ | |
| $= \displaystyle \left( \frac{2 y^{5/2}}{15} + \frac{2 y^{7/2}}{7} - \frac{13 y^4}{96} \right)\Bigg|_{y=0}^{y=4} $ $= \displaystyle \frac{216}{35}.$ |
This volume can be calculated by treating the region in the $x$-$y$ plane as either type I or II as seen in example 26.1.
In the following example, we see how it is sometimes necessary to change the order of integration in order to evaluate the integral.
For $\displaystyle \int \sin \left(y^2\right)dy,$ there is no anti-derivative in terms of elementary functions. 😥
Then we must change order of integration. That is, as a type II integral. In that case, we can easily compute the integral! 😃
For $\displaystyle \int \sin \left(y^2\right)dy,$ there is no anti-derivative in terms of elementary functions. 😥
Then we must change order of integration. That is, as a type II integral. In that case, we can easily compute the integral! 😃
| $\displaystyle \int_\limits{x=0}^{x=1} \left[ \int_{y=x}^{y=1} \sin \left(y^2\right) dy \right]dx$ | $ = \displaystyle \int_\limits{y=0}^{y=1} \left[ \int_{x=0}^{x=y} \sin \left(y^2\right) dx \right]dy $ |
| $= \displaystyle \int_\limits{y=0}^{y=1} \left[ \sin \left(y^2\right) x \Bigg|_{x=0}^{x=y} \right]dy $ | |
| $= \displaystyle \int_\limits{y=0}^{y=1} y\sin \left(y^2\right) dy $ $= \displaystyle \left( -\frac{1}{2} \cos \left(y^2\right) \right)\Bigg|_{y=0}^{y=1} $ | |
| $= \displaystyle \frac{1}{2} \big( 1 - \cos (1) \big).$ |