Double Integral evaluation help

In summary, because the integral is symmetrical around the y-axis, integrating by parts gives the same result.
  • #1
wilcofan3
27
0

Homework Statement



Evaluate the double integral [tex]\int \int_{R} ln(xy) dA[/tex] where [tex]R[/tex] is the rectangle bounded by [tex]x=e, x=e^2,y=1,y=e[/tex].

Homework Equations


[tex]ln (xy) = ln x + ln y[/tex]

The Attempt at a Solution



I was just wondering, do I need to do anything other than take the integral with respect to x, plug in the bounds, and then take the integral with respect to y, and plug in the bounds? I know to use integration by parts for ln x, and I know to break the ln xy up, but is it fine just to treat the ln y as a constant even if I do integration by parts? That would make it become x ln y. Thanks.
 
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  • #2
wilcofan3 said:

Homework Statement



Evaluate the double integral [tex]\int \int_{R} ln(xy) dA[/tex] where [tex]R[/tex] is the rectangle bounded by [tex]x=e, x=e^2,y=1,y=e[/tex].

Homework Equations


[tex]ln (xy) = ln x + ln y[/tex]

The Attempt at a Solution



I was just wondering, do I need to do anything other than take the integral with respect to x, plug in the bounds, and then take the integral with respect to y, and plug in the bounds? I know to use integration by parts for ln x, and I know to break the ln xy up, but is it fine just to treat the ln y as a constant even if I do integration by parts? That would make it become x ln y. Thanks.
If you're integrating with respect to x, then yes, ln y can be treated as a constant.

But what is "it" that would become x ln y?
 
  • #3
Mark44 said:
If you're integrating with respect to x, then yes, ln y can be treated as a constant.

But what is "it" that would become x ln y?

I was referring to the constant when I integrate with respect to x. Sorry, I was a little vague.

Thank you, though, I just was a little unsure on whether having a "u" substitution would change things up, but I should just treat it as a constant.

I ended up with the result [tex]2e^3 - 2e^2[/tex].
 
  • #4
wilcofan3 said:

Homework Statement



Evaluate the double integral [tex]\int \int_{R} ln(xy) dA[/tex] where [tex]R[/tex] is the rectangle bounded by [tex]x=e, x=e^2,y=1,y=e[/tex].

Homework Equations


[tex]ln (xy) = ln x + ln y[/tex]

The Attempt at a Solution



I was just wondering, do I need to do anything other than take the integral with respect to x, plug in the bounds, and then take the integral with respect to y, and plug in the bounds? I know to use integration by parts for ln x, and I know to break the ln xy up, but is it fine just to treat the ln y as a constant even if I do integration by parts? That would make it become x ln y. Thanks.

Just do the integral with respect to why then x, so it will be written dy dx,

for dy the bounds will just be from 1 to e,

then for dx it will go from e to e^2
 
  • #5
[tex]\int_{x=a}^b \int_{y= c}^d f(x)+ g(y) dy dx= \left(\int_{x= a}^b f(x)dx\right)\left(\int_{y= c}^d dy\right)+ \left(\int_{x=a}^b dx\right)\left(\int_{y= c}^d g(y)dy\right)[/tex]
[tex]= (d-c)\int_{x=a}^b f(x)dx+ (b-a)\int_{y= c}^d g(y)dy[/tex]
 

1. How do I evaluate a double integral?

Evaluating a double integral involves finding the area under a surface in two dimensions. To solve, you must first identify the limits of integration for both variables, then use a method such as Fubini's Theorem or change of variables to convert the double integral into two single integrals. Finally, solve each single integral separately and multiply the results together.

2. What is the purpose of a double integral?

A double integral has several purposes, including finding the volume under a surface, calculating the mass of a two-dimensional object, and determining the average value of a function over a given region.

3. What is Fubini's Theorem?

Fubini's Theorem states that for a double integral of a continuous function over a region, the integral can be computed by iterated integration, where the order of integration can be switched without changing the result. This theorem is useful in evaluating double integrals and can help simplify the process.

4. Can a double integral have a negative value?

Yes, a double integral can have a negative value. This can occur when the limits of integration are chosen in a way that results in the function being negative over the given region. It is important to carefully consider the bounds and orientation of the region when evaluating a double integral to ensure an accurate result.

5. Are there any shortcuts or tricks for evaluating double integrals?

There are several techniques that can be used to simplify the process of evaluating double integrals, such as symmetry, the use of polar coordinates, and the use of geometric properties to split the region of integration into smaller, simpler regions. However, these shortcuts may not always be applicable and it is important to have a solid understanding of the fundamental methods for evaluating double integrals.

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