Evaluate $\int\int(yx^2-2xy^2)dydx$: Get Help Here!

  • Thread starter der.physika
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In summary, the conversation discusses solving an integral with limits of 0 to 3 and -2 to 0 by splitting it into two integrals of yx^2 and 2xy^2. The final answer is 6.
  • #1
der.physika
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[tex]\int\int(yx^2-2xy^2)dydx[/tex]

limits for the first are 0 [tex]\longrightarrow[/tex] 3

limits for the second are -2 [tex]\longrightarrow[/tex] 0

solve! help
 
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  • #2
Hi der.physika! :smile:

(type "\int_0^3\int_{-2}^0" :wink:)

Just split it into two integrals, ∫∫ yx2 dydx and ∫∫ 2xy2 dydx …

what do you get? :smile:
 
  • #3
tiny-tim said:
Hi der.physika! :smile:

(type "\int_0^3\int_{-2}^0" :wink:)

Just split it into two integrals, ∫∫ yx2 dydx and ∫∫ 2xy2 dydx …

what do you get? :smile:

Okay so I took your advice and split the integral and I got 6, is that the correct answer?
 
  • #4
erm :redface: … i can't check your answer without seeing your calculations, can I? :wink:
 
  • #5
tiny-tim said:
erm :redface: … i can't check your answer without seeing your calculations, can I? :wink:

[tex]\int\int(yx^2dydx)-2\int\int(xy^2dydx)[/tex]

[tex]\int[\frac{1}{2}y^2x^2]=\int(-2x^2dx)=[\frac{-2}{3}x^3]=\frac{-54}{3}[/tex]

[tex]-2\int\int(xy^2dydx)=-2\int[\frac{1}{3}y^3x]=-2(-12)=24=\frac{72}{3}[/tex]

[tex]=\frac{72}{3}+\frac{-54}{3}=\frac{18}{3}=6[/tex]

is this okay?
 
Last edited:
  • #6
Sorry, I'm totally confused. :confused:

Where are your half-way integrals, ie after just one integration?

(and btw, which integral is going from 0 to 3, ∫ dx or ∫ dy?)
 
  • #7
tiny-tim said:
Sorry, I'm totally confused. :confused:

Where are your half-way integrals, ie after just one integration?

(and btw, which integral is going from 0 to 3, ∫ dx or ∫ dy?)

How do you put in limits on the integral? I don't know how to put the code into LaTex
 
  • #8
I showed you above … type "\int_0^3" and "\int_{-2}^0"

(you need {} if the limit has more than one character, eg -2)
 

1. What is the purpose of evaluating a double integral?

Evaluating a double integral allows us to calculate the volume under a three-dimensional surface or the area of a two-dimensional region on a plane. This is useful in many applications, such as physics, engineering, and economics.

2. How do you evaluate a double integral?

To evaluate a double integral, we use the process of iterated integration. This involves integrating the inner function with respect to one variable, and then integrating the resulting function with respect to the other variable. We can also use various techniques such as substitution or integration by parts to simplify the integral.

3. What is the function being integrated in this specific double integral?

The function being integrated in this double integral is (yx^2-2xy^2). This is a multivariable function with two variables, y and x.

4. Can this double integral be solved analytically?

Yes, this double integral can be solved analytically using the process of iterated integration. However, depending on the complexity of the function, it may be challenging to solve by hand and require the use of computer software or numerical methods.

5. What are the possible benefits of using a double integral to evaluate this function?

Using a double integral allows us to accurately calculate the volume or area of a complex shape, which would be difficult or impossible to do with a single integral. It also allows us to solve problems involving multivariable functions, which are common in many fields of science and engineering.

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