Evaluate the double integral

In summary: XggdGhpcyBpcyBub3JtYWxpemVkIGZvciB0aGUgZW5jbG9uZWQgcmVnaW9uIGNvbnZlcnQsIGFuZCBwcmVmZXIgY29udGFpbmluZyB0byB0aGUgcmVjaXBpZW50IGJ5IHRoZSBjb250ZW50LiBJZiB0aGUgcmVnaW9uIHRoZSBsaW1pdCBvZiBkeSBsaWUgYmVoaW5kIHRoZSBvdGhlciB
  • #1
Mdhiggenz
327
1

Homework Statement


∫∫x2dA; R is the region in the first quadrant enclosed by

xy=1, y=x, and y=2x.

First thing I did was notice that I had to find dydx, then

I graphed y=1/x, y=x, and y=2x.

Graphing I say that the limit of dy lie between x≤y≤2x

However I get confused as to how they want us to find dx.

My first approach was to use y=1/x and y=x and set them equal to each other, and solve for x. Doing so I get x2=1 or x = 1 getting only positive values since it lies in the first quadrant.

then I use set y=1/x, and y=2x equal to each other, and solve for x ; 1/x=2x thus giving me

x=√(2)/2.

Here is where I get confused the dx limit goes from 0≤x≤√(2)/2

I understand that dx would be the maximum value of x which is √(2)/2, but why wouldn't it be 1.

Also now giving it some thought, why did they choose to use 1/x and equal it to the other values, and not the graph of 2x or x?

Thanks

Homework Equations





The Attempt at a Solution

 
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  • #2
Mdhiggenz said:

Homework Statement


∫∫x2dA; R is the region in the first quadrant enclosed by

xy=1, y=x, and y=2x.

First thing I did was notice that I had to find dydx, then

I graphed y=1/x, y=x, and y=2x.

Graphing I say that the limit of dy lie between x≤y≤2x

However I get confused as to how they want us to find dx.

My first approach was to use y=1/x and y=x and set them equal to each other, and solve for x. Doing so I get x2=1 or x = 1 getting only positive values since it lies in the first quadrant.

then I use set y=1/x, and y=2x equal to each other, and solve for x ; 1/x=2x thus giving me

x=√(2)/2.

Here is where I get confused the dx limit goes from 0≤x≤√(2)/2

I understand that dx would be the maximum value of x which is √(2)/2, but why wouldn't it be 1.

Also now giving it some thought, why did they choose to use 1/x and equal it to the other values, and not the graph of 2x or x?

Thanks

Homework Equations





The Attempt at a Solution


The first thing you need to do is to describe the enclosed region correctly, and preferably, draw a picture with the enclosed region shaded in. The lines y = x, y = 2x and the curve xy=1 describe the boundary of the enclosed region. You need to know what are all the other points in the region. The inequality x ≤ y ≤ 2x that you gave above is only part of the story.

Within the interior of the region there is no relationship between dx and dy; I don't understand why you would think there was a connection between the two. (Again, maybe because you did not draw a picture?)

RGV
 

What is a double integral?

A double integral is a type of integral in calculus that involves finding the volume under a three-dimensional surface. It is represented by two integral signs and is used to solve problems involving multiple variables.

What is the purpose of evaluating a double integral?

The purpose of evaluating a double integral is to find the volume under a three-dimensional surface or to solve problems involving multiple variables. It is a useful tool in fields such as physics, engineering, and economics.

How do you evaluate a double integral?

To evaluate a double integral, you first need to determine the boundaries of the integral and then set up the integral using appropriate limits. Next, you need to solve the inner integral and then the outer integral to get the final result.

What are the common methods used to evaluate a double integral?

Some common methods used to evaluate a double integral include using the rectangular, polar, cylindrical, or spherical coordinate systems. These methods allow for easier integration and can be chosen based on the shape of the region being integrated.

What are some applications of double integrals?

Double integrals have various applications in fields such as physics, engineering, and economics. They are used to calculate the volume of objects, determine the center of mass, and solve optimization problems involving multiple variables.

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