Double Integral: Evaluating $II_{5a}$ in $R=[0,2] \times [-1,1]$

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Discussion Overview

The discussion revolves around evaluating the double integral \( II_{5a} \) over the region \( R = [0,2] \times [-1,1] \) for the function \( xy\sqrt{x^2+y^2} \). Participants explore the implications of the function's symmetry and the limits of integration.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes evaluating the double integral \( II_{5a} = \iint\limits_{R} xy\sqrt{x^2+y^2}\, dA \) over the specified region.
  • Another participant notes that the function \( f(x,y) = xy\sqrt{x^2+y^2} \) is odd in \( y \) and concludes that the integral over \( y \) from -1 to 1 results in zero, suggesting \( II_{5a} = 0 \).
  • Some participants express concern about a potential typo in the limits of integration, specifically questioning the upper limit of \( x \) being 2 instead of 1.
  • One participant argues that regardless of whether the limit is 1 or 2, the \( y \)-integral remains zero, leading to the conclusion that the double integral is still zero.

Areas of Agreement / Disagreement

There is disagreement regarding the limits of integration, with some participants questioning a potential typo. However, there is a consensus that the double integral evaluates to zero based on the symmetry of the function.

Contextual Notes

The discussion includes uncertainty about the correct limits of integration and the implications of the function's symmetry on the evaluation of the integral.

karush
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$\textsf{a. Evaluate :}$
\begin{align*}\displaystyle
R&=[0,2] \times [-1,1]\\
II_{5a}&=\iint\limits_{R}xy\sqrt{x^2+y^2}\, dA
\end{align*}
next step?
$$\displaystyle\int_0^1 \int_{-1}^1
xy\sqrt{x^2+y^2}\, dxdy$$
 
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Re: 15.2.a dbl int

karush said:
$\textsf{a. Evaluate :}$
\begin{align*}\displaystyle
R&=[0,2] \times [-1,1]\\
II_{5a}&=\iint\limits_{R}xy\sqrt{x^2+y^2}\, dA
\end{align*}
next step?
$$\displaystyle\int_0^1 \int_{-1}^1
xy\sqrt{x^2+y^2}\, dxdy$$
If $f(x,y) = xy\sqrt{x^2+y^2}$ then $f(x,-y) = -f(x,y)$ and therefore $$\int_{-1}^1f(x,y)\,dy = 0.$$ So $$II_{5a} = 0.$$
 
Re: 15.2.a dbl int

wait
one of the limits is [0,2]
looks like a typo in the Integral
 
Re: 15.2.a dbl int

karush said:
wait
one of the limits is [0,2]
looks like a typo in the Integral
Makes no difference whether it is 1 or 2. The $y$-integral is zero and therefore the double integral is zero.
 

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