Very tricky double integration problem

[jcook735]

Calculate the double integral.

∫∫((5x)/(xsquared)+(ysquared)) dA, R = [1,4] x [0,1].
Find the value of the integral.


Well I tried the integral starting with x and then starting with y and couldn't do it either way. Ill show the method using x first. To start off I tried using U-substitution to be able to integrate according to y. I had u = x^2 and du=2xdx. I pulled the 5 out of the equation, then put u in place of x^2 squared at the bottom of the equation, then put a 2 in the equation and a .5 outside the equation so i could replace 2xdx with du. So i ended up with the equation (5/2)∫du/(u+y^2). I then took the integral so that I ended up with (5/2) ln|u+y^2| (evaluated from 1 to 4). I then plugged in x^2 for u. So then I plugged in 4 and 1, and got ∫{(5/2)[ln|16+y^2| - ln|1+y^2|}dy. So then I set it up for the integral according to y. I used integration by parts on ln|16+y^2|, and got stuck. I found u=ln|16+y^2| and dv to equal dy. Then I found du to equal (2y)/(16+y^2) and v to equal y. I then plugged this into the given formula, vu-∫vdu. This is where I got stuck. How do I take the integral of vdu, aka (2y^2)/(16+y^2)? Help would be greatly appreciated.

 

[tiny-tim]

hi jcook735!

(try using the X² icon just above the Reply box )

How do I take the integral of vdu, aka (2y^2)/(16+y^2)? Help would be greatly appreciated.

that looks pretty good so far!

ok, you need to integrate 2y²/(16+y²)

just write it 2 - 32/(16+y²)

(the second part is a trig integral)

 

[Mark44]

For your last question, 2y^2/(16 + y^2) = 2 - 32/(y^2 + 16)

 

[jcook735]

oh haha my bad I didnt know you could do that. why does y^2 turn into 32??

 

[jcook735]

OH! I got it. Thank you guys so much :)

 

[jcook735]

can yall check to make sure i did my integration correctly? for the equation you gave me, I should end up with 2y - (1/2)arctan(y/4), correct?

 

[tiny-tim]

can yall check to make sure i did my integration correctly? for the equation you gave me, I should end up with 2y - (1/2)arctan(y/4), correct?

You need to write it out more clearly if you want people to check it.

But anyway, wasn't there an integral of something over (1 + y²) also?

 

[Mark44]

Also, since this was a definite integral, you should end up with a number.

tiny-tim - That integral of something over 1 + y^2 was probably what turned into the arctan part.

 

[jcook735]

alright my bad. I am taking the ∫2- 32/16+y².

first, i separated the parts, so I have ∫2 - ∫(32)/(16+y²).

The first integral is 2y.

For the second integral, i pulled 16 out of the bottom, so i had (32)/((16)(1+(y/4)²)). I canceled out the 32 and 16, so i ended up with (2)/(1+(y/4)²). So then its 2∫(1)/(1+(y/4)²). This should turn out to be (8)arctan(y/4), correct?

So the whole equation should be 2y-(8)arctan(y/4).

 

[tiny-tim]

That looks ok, but it was the whole thing that was bothering me …

you had " ∫{(5/2)[ln|16+y^2| - ln|1+y^2|}dy " in your first post, and the second bit seems to have gone missing.

 

[jcook735]

Oh well I just needed help on the first part cause the second part followed suit. I got it right thank you both very much!

 

[._|evo|_.]

Cool nice job!