Reverse the order of the integration in the following expression \int_{-1}^0 \int_0^{8+2x} Q dy...

Question:

Reverse the order of the integration in the following expression

{eq}\displaystyle \int_{-1}^0 \int_0^{8+2x} Q dy dx +\int_0^4 \int_0^{8-2x}Q dy dx {/eq}

where Q is an unknown function of x and y

Evaluating Double Integrals using Iterated Integrals:

Consider the integral of the function {eq}F(x,y) {/eq} over the planar region {eq}D {/eq} on the xy-plane, expressed as:

$$\iint \limits_D F(x,y) \,dA $$

Suppose the region of integration is a Type I region, one which is bounded by continuous functions of x, such that {eq}D = \left\{\, (x,y) \,|\, a \le x \le b \text{ and } g_1(x) \le y \le g_2(x) \,\right\} {/eq}.

Then, the double integrals above can be written as the iterated integrals:

$$\iint \limits_D F(x,y) \,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} F(x,y) \,dy \,dx $$

On the other hand, suppose the region of integration is a Type II region, one which is bounded instead by continuous functions of y, such that {eq}D = \left\{\, (x,y) \,|\, c \le y \le d \text{ and } h_1(y) \le x \le h_2(y) \,\right\} {/eq}.

Then, the double integrals above can also be written as the iterated integrals:

$$\iint \limits_D F(x,y) \,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} F(x,y) \,dx \,dy $$

Answer and Explanation:

Given the sum of the two iterated integrals {eq}\begin{align*} \int_{-1}^0 \int_0^{8 + 2x} Q(x,y) \,dy \,dx + \int_0^4 \int_0^{8 - 2x} Q(x,y) \,dy \,dx \end{align*} {/eq}

Let {eq}R1 \text{ and } R2 {/eq} be the regions of integration of the first and second iterated integrals, respectively. Looking at the order of integration, {eq}dy \,dx {/eq}, these two regions are defined as Type I regions.

Looking at the limits of integration:

  • Region {eq}R1 {/eq} is bounded above by {eq}y = 8 + 2x {/eq} and below by {eq}y = 0 {/eq} on the interval {eq}-1 \le x \le 0 {/eq}; and,
  • Region {eq}R2 {/eq} is bounded above by {eq}y = 8 - 2x {/eq} and below by {eq}y = 0 {/eq} on the interval {eq}0 \le x \le 4 {/eq}.

Graph of Type I Regions R1 and R2

Graph of Type I Regions R1 and R2

To reverse the order of integration, the union of the two regions {eq}R1 \text{ and } R2 {/eq}, which represents the whole region of integration, must be defined as Type II region/s.

This can be done by dividing the region of integration into two smaller regions using the horizontal line {eq}y = 6 {/eq}. Let {eq}R3 \text{ and } R4 {/eq} be the resulting smaller Type II regions.

Convert the equations of the diagonal lines into functions of y:

{eq}\begin{align*} &y = 8 + 2x \\ &\Rightarrow 2x = y - 8 \\ &\Rightarrow x = \frac{y}{2} - 4 & \text{[Left Diagonal Line (green)]} \\ \\ &y = 8 - 2x \\ &\Rightarrow 2x = 8 - y \\ &\Rightarrow x = - \frac{y}{2} + 4 & \text{[Right Diagonal Line (red)]} \end{align*} {/eq}


Then, based on the graph and the functions above, the two Type II regions can be defined as {eq}R3 = \left\{\, (x,y) \,|\, 0 \le y \le 6 \text{ and } -1 \le x \le - \frac{y}{2} + 4 \,\right\} \text{ and } R4 = \left\{\, (x,y) \,|\, 6 \le y \le 8 \text{ and } \frac{y}{2} - 4 \le x \le - \frac{y}{2} + 4 \,\right\} {/eq}.

Graph of Type II Regions R3 and R4

Graph of Type II Regions R3 and R4

Therefore, the given iterated integrals when the order of integration is reversed are equal to:

{eq}\begin{align*} &\int_{-1}^0 \int_0^{8 + 2x} Q(x,y) \,dy \,dx + \int_0^4 \int_0^{8 - 2x} Q(x,y) \,dy \,dx = \iint \limits_{R3} Q(x,y) \,dA + \iint \limits_{R4} Q(x,y) \,dA \\ &\boxed{ \int_{-1}^0 \int_0^{8 + 2x} Q(x,y) \,dy \,dx + \int_0^4 \int_0^{8 - 2x} Q(x,y) \,dy \,dx = \int_0^6 \int_{-1}^{-\frac{y}{2}+4} Q(x,y) \,dx \,dy + \int_6^8 \int_{\frac{y}{2}-4}^{-\frac{y}{2}+4} Q(x,y) \,dx \,dy } \end{align*} {/eq}


Learn more about this topic:

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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
512

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