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

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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