MHB 232.5a Evaluate the double integral

[karush]

$\tiny{232.5a}\\
\textsf{Evaluate the double integral}$
\begin{align*}\displaystyle
I_a&=\iint\limits_{R} xy\sqrt{x^2+y^2} \, dA \\
R&=[0,2]\times[-1,1]
\end{align*}
Ok, just want to see if I made the first step correct.
this looks like simply a rectangle so x and y are basically interchangeable
\begin{align*}
&=\int_{-1}^{1} \int_{0}^{2} xy\sqrt{x^2+y^2} \,dx \,dy
\end{align*}
however the next step looks kinda daunting

 

[Greg]

How about the substitution $x=y\tan\theta$?

 

[karush]

How about the substitution $x=y\tan\theta$?

are you suggesting that

\begin{align*}
&=\int_{-1}^{1} \int_{a}^{b} y^2\tan\theta \, \sqrt{y^2\tan^2\theta+y^2} \,d\theta \,dy
\end{align*}

not sure what a and b would be on the limits

 

[tkhunny]

$\tiny{232.5a}\\
\textsf{Evaluate the double integral}$
\begin{align*}\displaystyle
I_a&=\iint\limits_{R} xy\sqrt{x^2+y^2} \, dA \\
R&=[0,2]\times[-1,1]
\end{align*}
Ok, just want to see if I made the first step correct.
this looks like simply a rectangle so x and y are basically interchangeable
\begin{align*}
&=\int_{-1}^{1} \int_{0}^{2} xy\sqrt{x^2+y^2} \,dx \,dy
\end{align*}
however the next step looks kinda daunting

Ummm... No. Do an experiment for me. Take x = 2 and evaluate the single integral. Then, back up and take x = 1 and evaluate the single integral. See if you notice something. This is an eyeball problem.

Definition: "Eyeball Problem" - Just look at it and call out the answer before your neighbor beats you to it.

 

[karush]

that is giving me a headache

I presume you mean separate x and y

 

[tkhunny]

that is giving me a headache

I presume you mean separate x and y

No, I mean just pick x = 2 and then ignore that x exists and ignore the inner integral.

$\int\limits_{-1}^{1}2y\sqrt{4+y^2}\;dy$

What do you get?

Now, x = 1.

$\int\limits_{-1}^{1}y\sqrt{1+y^2}\;dy$

What do you get?