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

[karush]

$\textsf{a. Evaluate :}$
\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?
$$\displaystyle\int_0^1 \int_{-1}^1
xy\sqrt{x^2+y^2}\, dxdy$$

 

[Opalg]

Re: 15.2.a dbl int

$\textsf{a. Evaluate :}$
\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 \(\displaystyle \int_{-1}^1f(x,y)\,dy = 0.\) So \(\displaystyle II_{5a} = 0.\)

 

[karush]

Re: 15.2.a dbl int

wait
one of the limits is [0,2]
looks like a typo in the Integral

 

[Opalg]

Re: 15.2.a dbl int

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.