Question

Basic multiplication !!! ln(1+r^2)rdrd(theta)

So i have to multiply ln(1+r^2) by r

dont want to make silly mistakes....

Answer 1

Are you sure this is multiplications? This looks more like integration from calculus with multivariables

Answer 2

It a double integral question with polar coordinates\[\int\limits_{0}^{5}\int\limits_{0}^{\sqrt{25-x^{2}}}\ln \left( 1+x^{2}+y^{2} \right)dydx\]

Answer 3

and i ended up with what i posted when trying to change to polar coordinates.... does that make sense ? hope im not doing it wrong :/

Answer 4

So in your second, post that is the original equation? and when you switched it to polar coordinates you got the equation in your opening post?

Answer 5

Yes thats correct , with the limits being
\[\int\limits_{0}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{5}\]

Answer 6

First one is pi/2 if its too small to see :)

Answer 7

So do i just take r to the front

\[rln \left( 1+r^2{?} \right)\]

Answer 8

and then intergrate ?

Answer 9

Give me a second, I haven't done these problems in the longest time. lol. If someone else comes by before me they can give it a shot, trying to relearn this real quick

Answer 10

No worries :) i appreciate your help :)

Answer 11

Have you ever heard of "u" substitution? Let u = 1 + r^2

Answer 12

Yes i have

Answer 13

Do you have the answer from teh back of the book or something i wanna check something

Answer 14

No i dont sorry :(

Answer 15

okay, so you know to work from the inside out from your polar coordinate double integral equation, do the "u" sub I told you, since we can't multiply "r" into ln(1+r^2).

Answer 16

so we get....
rln(u)

where u =1+r^2
du/dr=2r
dr=du/2r

????

Answer 17

Yeah. The odd thing is, I'm getting a negative answer which is incorrect because integration over double integrals usually provide and area with positive value not negative. interesting. are you sure the limits are correct?

Answer 18

hmmm.. they could be wrong
ill show you what i did

Answer 19

\[y^{2}\le \sqrt{25-x^{2}}\]\[x^{2}+y^{2}\le25\]\[x \ge0 , x \le5\]

Answer 20

okay, that works for the limits of "r" what about "theta"

Answer 21

I havent got that far yet... what answer did u get once you intergrated it??

Answer 22

Final Answer as\[\frac{ -25 }{ 52 }\frac{ \pi }{ 2 }\]

Answer 23

Sorry i havent got that far yet...im still trying to intergrate with sub !! its been a while since i have done it

Answer 24

Yeah "u" subs will come back and haunt you. but let me ask someone real quick before you start going with the "u" subs

Answer 25

@ganeshie8 Can you help us a bit here ganeshie8

Answer 26

I agree!! parts all the way!!

Answer 27

@wio Need your expertise real quick, second pair of eyes please.

Answer 28

I typed the function into wolf frame alpha and got a circle that had its volume under the xy plane, maybe that's why it's negative, but we'll see from the other experts here.

Answer 29

Hmm so does that mean my limits are wrong ?

Answer 30

No. Sometimes the answers are negative. Is this from your teacher or a textbook?

Answer 31

https://www.google.com/#q=graph+ln%281%2Bx^%282%29%2By^%282%29%29

that's what it looks like.

Answer 32

its a question of a previous years exam, but they haven't provided solutions to the problem

Answer 33

Oh. Interesting. Well, it's best to wait for another pair of eyes to see this before you go through, if else, skip this one and study another thing while you wait for confirmation on this. @Directrix

Sorry I couldn't fully help you.

Answer 34

Yes plenty of other things to do.... dont suppose your interested in a eigenvalue and eigenvector problem??

Answer 35

Thanks for your help !! it got me thinking more having someone to talk to !!

Answer 36

You're welcome. I'm going to sleep now it's late over here. Good night.

Answer 37

Okay then. Thank you once again!!!