Question

Need some help with this double integral problem.

Answer 1

\[\int\limits_{0}^{1}\int\limits_{0}^{4}xy \sqrt{x^2+y^2}dydx\]

Answer 2

\[\int\limits_{0}^{4}xy \sqrt{x^2+y^2}dy\]

Answer 3

@agent0smith How would you go about this problem from here?

Answer 4

Hmm, you might need to try polar coordinates?

Answer 5

I am not too familiar with them. I have been working on this problem for more than an hour. I can't seem to figure it out.

Answer 6

\You might want to google them but here's some basics: http://www.mathwords.com/p/p_assets/p78.gif
I haven't done double integrals in a while, but seems kinda hard to do it in its current form
\[\large \int\limits\limits_{}^{}\int\limits\limits_{}^{}xy \sqrt{x^2+y^2}dydx = \]\[\large \int\limits\limits_{}^{}\int\limits\limits_{}^{}r \cos \theta * r \sin \theta * r dr d \theta = \]\[\large \int\limits\limits\limits_{}^{}\int\limits\limits\limits_{}^{}r^3 \sin \theta \cos \theta dr d \theta = \] you can also apply a trig identity here \[\large \frac{ 1 }{ 2 } \int\limits\limits\limits\limits_{}^{}\int\limits\limits\limits\limits_{}^{}r^3 \sin 2\theta dr d \theta \]

Answer 7

I will do some reading up on them. Thanks for the help!

Answer 8

Angel Smite, did you miss an r maybe?
\[\large\rm \int\limits\int\limits xy\sqrt{x^2+y^2}~dy~dx\]\[\large\rm \int\limits\int\limits(r \cos \theta)(r \sin \theta)\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}~r~dr~d \theta\]

Answer 9

http://tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx

Yeah there should be the extra r for the dy dx replacement

Answer 10

So it should be \[\large \frac{ 1 }{ 2 } \int\limits\limits\limits\limits\limits_{}^{}\int\limits\limits\limits\limits\limits_{}^{}r^4 \sin 2\theta dr d \theta\]
But idk if i'm willing to try to relearn how to finish this

Answer 11

lol :D

Answer 12

How about the limits?

Answer 13

@Loser66 did you not read the last line of my last post?

Answer 14

Hmm you could just do this one straight up in rectangular coordinates if you want.\[\large\rm \int\limits_{x=0}^1~\color{orangered}{\int\limits_{y=0}^4 xy\sqrt{x^2+y^2}~dy}dx\]It's quite a bit of algebra, but nothing too difficult.

Answer 15

If you want to do a u-sub, fine and dandy.
But you should try to get comfortable with these types of things from Calc 2.\[\large\rm \int\limits y\sqrt{a+y^2}dy\approx \frac23(a+y^2)^{3/2}\]Just sort of think ahead like this.

Answer 16

And then you think, oh but we're missing the 2y when the y^2 gets differentiated, so we need a 1/2 to compensate.\[\large\rm \int\limits y\sqrt{a+y^2}~dy=\frac12\frac23(a+y^2)^{3/2}\]

Answer 17

If that's too confusing... fine... u-sub it. :p
Anyway, you treat the x as constant while you're doing all this business in y.\[\large\rm \int\limits\limits\limits_{x=0}^1~\color{orangered}{\left(x\frac12\frac23(x^2+y^2)^{3/2}\right)_{y=0}^4}dx\]

Answer 18

Pull the fraction out front,
Then evaluate this at your y boundaries,\[\large\rm \frac13\int\limits_{x=0}^4 x\color{orangered}{\left[(x^2+4^2)^{3/2}-(x^2+0^2)^{3/2}\right]}~dx\]Then distribute the x and integrate each piece separately.
The first piece will be exactly like the y's were,
the other piece will be power rule.

Answer 19

Just another option to consider :)

Answer 20

I did, but didn't see the limits.