Question

Express D as a region of type II. $$D = \{(x, y) | 0 \le y \le x, y \le x \le 3\}$$ $$D = \{(x, y) | 0 \le y \le x, 0 \le x \le y\}$$ $$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le y\}$$ $$D = \{(x, y) | 0 \le y \le 3, y \le x \le 3\}$$ $$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le 3\}$$ Evaluate the double integral in two ways.

          Express D as a region of type II.
$$D = \{(x, y) | 0 \le y \le x, y \le x \le 3\}$$
$$D = \{(x, y) | 0 \le y \le x, 0 \le x \le y\}$$
$$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le y\}$$
$$D = \{(x, y) | 0 \le y \le 3, y \le x \le 3\}$$
$$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le 3\}$$
Evaluate the double integral in two ways.

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express d as a region of type ii d x y 0 le y le x y le x le 3 d x y 0 le y le x 0 le x le y d x y 0 le y le 3 0 le x le y d x y 0 le y le 3 y le x le 3 d x y 0 le y le 3 0 le x le 3 evaluat 89489

Added by Aaron B.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Shaiju T

04/11/2022

Step 1

A region of type II is defined by inequalities of the form $D = \{(x, y) | c \le y \le d, h_1(y) \le x \le h_2(y)\}$. This means that y is bounded by constants, and x is bounded by functions of y. Let's analyze the given options. We are looking for an option Show more…

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Express D as a region of type II. $$D = \{(x, y) | 0 \le y \le x, y \le x \le 3\}$$ $$D = \{(x, y) | 0 \le y \le x, 0 \le x \le y\}$$ $$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le y\}$$ $$D = \{(x, y) | 0 \le y \le 3, y \le x \le 3\}$$ $$D = \{(x, y) | 0 \le y \le 3, 0 \le x \le 3\}$$ Evaluate the double integral in two ways.

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00:01 In this question, we are given double integration, that is integration and then integration on d, x, da, and where d is included region by the lines, y equal to x, y equal to 0, and x equal to 3.

00:14 And we have to express d as a region of type 1 and type 2.

00:19 So, first of all, we will express d as a region of type 1.

00:26 So for type 1, first of all, we will make the graph here.

00:29 So, let us suppose this is our x -axis and this is our y -axis.

00:35 So, y -equal to x will be this line.

00:38 Y -equal to 0 will be this line.

00:40 And x -equal to 3 will be this line.

00:43 Let us say.

00:45 So, we have to find double integration on this area.

00:52 So now, we will find area.

00:56 So, in type 1, we will take a vertical strip with length, dx, and it will move from 0 to and its upper part will be on y equal to x -axis.

01:14 So for type 1, d can be written as x -coma -y such that x belongs to 0 to 3 and y belongs from 0 to x.

01:28 So if we see the option, then clearly option d will be correct option here.

01:35 Now, we will define d as type 2.

01:43 So for type 2, we will also make graph again.

01:48 So let us say this is our y -axis, this is our x equal to 3 line.

01:54 And now in type 2, we will take this horizontal type of strip.

02:00 So let's this thickness is d -y...

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