Integrate Iterated Integrals: Solving Homework

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It's always good to double check your work using different methods to catch any mistakes. Great job!In summary, the conversation was about evaluating a double integral and the attempt at solving it. The correct answer was found to be 14, with the help of double checking the work using different methods.
  • #1
-EquinoX-
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Homework Statement



I need to evaluate this
[tex] \int_{-2}^{0} \! \int_{-\sqrt{9-x^2}}^{0} 2xy\,dy\,dx. [/tex]

Homework Equations


The Attempt at a Solution



So I first tried integrating with respect to dy and I have:

[tex] xy^2 [/tex] and evaluate that from [tex] -\sqrt{9-x^2}[/tex] to 0 , I will have [tex] -18x + 2x^3[/tex]. Then I will integrate this again over dx, then I have [tex] 9x^2 + \frac{x^4}{2} [/tex] and I evaluate this from -2 to 0. which results in -44. Where did I do wrong?
 
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  • #2
-EquinoX- said:

Homework Statement



I need to evaluate this
[tex] \int_{-2}^{0} \! \int_{-\sqrt{9-x^2}}^{0} 2xy\,dy\,dx. [/tex]


Homework Equations





The Attempt at a Solution



So I first tried integrating with respect to dy and I have:

[tex] xy^2 [/tex]

and evaluate that from [tex] -\sqrt{9-x^2}[/tex] to 0 , I will have [tex] -18x + 2x^3[/tex].
You should have gotten -x(9 - x^2) here, or x^3 - 9x. You're off by a factor of 2.
-EquinoX- said:
Then I will integrate this again over dx, then I have [tex] 9x^2 + \frac{x^4}{2} [/tex] and I evaluate this from -2 to 0. which results in -44. Where did I do wrong?
It looks like you made another mistake, this time a sign error. From your previous work, you should have gotten -9x^2 + 1/2 * x^4.

The region over which integration takes place is a portion of the southwest quadrant of a circle of radius 3, centered at (0, 0). The region is bounded by the axes, the said circle, and the line x = -2.

Just to check my work, I integrated over the same sized region in the northeast quadrant of the circle, this time bounded by the line x = 2. You can't do this in general, but the integrand in this problem, 2xy, is unaffected by changes in signs to both variables. Both integrations resulted in values of 14, but the second way was simpler, with fewer chances to make sign errors.
 
  • #3
So I should get 14 as a final result if I did everything correctly?
 
  • #4
Unless I made a mistake doing the problem in two different ways...
 
  • #5
Yes, 14 is correct.
 

What is an iterated integral?

An iterated integral is a type of integral that involves solving a function with multiple variables by integrating one variable at a time. It is commonly used in higher level math and physics to solve complex problems.

Why is it important to learn how to integrate iterated integrals?

Integrating iterated integrals is essential in many scientific and mathematical fields, such as physics, engineering, and economics. It allows for solving complex problems and obtaining accurate results.

What are the steps for solving an iterated integral?

The first step is to determine the limits of integration for each variable. Then, integrate the innermost variable and substitute the limits of integration. Next, integrate the resulting function with respect to the next variable and again substitute the limits of integration. Repeat this process until all variables have been integrated.

What are some common mistakes to avoid when solving iterated integrals?

One common mistake is forgetting to substitute the limits of integration for each variable after integrating. Another mistake is integrating the variables in the wrong order, which can lead to incorrect results.

How can I improve my skills in solving iterated integrals?

Practice is key to improving your skills in solving iterated integrals. Start with simpler problems and gradually work towards more complex ones. It can also be helpful to review the fundamental principles of integration and familiarize yourself with common techniques and formulas.

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