- anonymous

int(int(sqrt(x^2+y^2), x = y .. sqrt(8*y)), y = 0 .. 8)
Please help me! :( this should be calculated using polar cor

- schrodinger

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- Owlfred

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- amistre64

double inting on polars eh .... sounds delightful :)

- amistre64

{SS} sqrt(x^2 +y^2) dr.dt

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## More answers

- amistre64

be right back .....

- amistre64

y <= x <= sqrt(8y)
0 <= y <= 8

- amistre64

{SS} sqrt(x^2 +y^2) dx.dy seems more appropriate than dr.dt
we gotta convert this to polar?

- anonymous

yeah we need to convert it to polar,,what i dnt understand is..How I should change my boundaries

- amistre64

id have to draw a picture..... :)

- amistre64

i know the cap is a sphere ... but for the domain I need to see it

- anonymous

This is the question though

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- amistre64

nice :)

- amistre64

this is my picture

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- amistre64

as we sweep this across our dt doesnt really change does it?it goes for pi/4 to pi/2 if anything right?

- amistre64

and our radius goes from 0 to the longest distance there is along the y=x line; or 8sqrt(2)

- anonymous

yeah I dont understand what my radius should be..

- amistre64

if i see it right; your radius goes from 0 to 8sqrt(2)

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- anonymous

how do u say 8sqart2..i dnt understand that part

- amistre64

your radius is a constant change; your theta depends on sqrt(8y) right?

- amistre64

ok... the radius is the distance from the origin to the end of the line right?
it travels up the y=x line; which is a 45 degree angle
the total length of that line is from 0, the origin, up to (8,8) the common point

- amistre64

the slant of a rt tri with 45 degrees is sqrt(2); this goes 8 times longer...
so: 0 <= radius <= 8sqrt(2)

- amistre64

pi/4 <= theta <= sqrt(8y), just gotta figure out if that sqrt(8y) need to be altered...

- anonymous

hello anthony sir,,,how are you ??????you are excellent.

- amistre64

howdy sath :) im fine thnx

- anonymous

i wish i were there with you!

- amistre64

you just bored? or bad times?

- anonymous

nah...but am great of yours.wanna learn something from you.

- anonymous

atleast way to study.

- anonymous

im trying to calculate it and I got 256pi/3...but it turns out to be wrong :(

- amistre64

{SS} f(x,y) dA <=> {SS} f(rcos, rsin)r dr.dt

- anonymous

so i have to integrate this right
int(int(r^2, r = 0 .. 8*sqrt(2)), theta = (1/4)*Pi .. (1/2)*Pi)

- amistre64

i think the upper limit for the theta is spose to correspond to the x = sqrt(8y) function somehow

- amistre64

x = r cos(t)
y = r sin(r) right
r cos(t) = sqrt(8y)
r cos(t) = sqrt(8 rsin(t))

- amistre64

r^2 cos^2(t) = 8r sin(t) solve for 't' i think

- amistre64

or solve for r :)

- amistre64

finding 't' sounds more reasonable to me; unless we can convert around to make it simpler

- anonymous

this is sooo difficult.. :(

- amistre64

well the good news is that the equation sqrt(x^2 +y^2) is just the unit sphere :) and should be easy to convert to polars

- anonymous

yeah..as I said earlier after I used those boundaries (theetha bound as pi/4to pi/2 and r bound 0 to \[8\sqrt{2}\] the answer I got was 256pi/3 but its wrong

- amistre64

r^2 cos^2(t) = 8r sin(t) ; /r
r cos^2(t) = 8 sin(t) ; indentity
r (1-sin^2(t)) = 8 sin(t)
r - r sin^2(t) = 8 sin(t)......
it gonna be wrong because you sweep 90 degrees each time thru.
You have to only sweep out the are from p/4 to the curve x=sqrt(8y)

- amistre64

you are ending up taking the area 1/16th of a sphere if you do pi/4 to pi/2. and we dont want the whole area...

- amistre64

does that make sense?

- anonymous

so is the answer pi/16?

- amistre64

i got no clue what the answer is yet :)
I could check it with the x and y stuff, but the polar version I aint got ...

- anonymous

yeah please...

- amistre64

hold on...

- anonymous

sure

- amistre64

well, i had to seek a higher source lol

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- amistre64

http://www.wolframalpha.com/widgets/gallery/view.jsp?id=f5f3cbf14f4f5d6d2085bf2d0fb76e8a

- anonymous

but still it turns out to be incorrect..

- amistre64

is it looking for an exact? or an approximation?

- anonymous

I think we should keep the answers in terms of pi..

- amistre64

i aint having no luck at it....

- anonymous

its fine then..thanks a lot :)

- anonymous

i have the answer exactly if you want.
1024/45+(1024/45)*sqrt(2)

- anonymous

and i have the correct integral. I used maple to solve though. I can try to do it by hand if you want.

- anonymous

\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]

- anonymous

can you read that?

- anonymous

ya i can read it..

- anonymous

omg its crct....pls tell me hw u do it

- anonymous

i have it down to this:
\[8^3/6*\int\limits_{0}^{\pi/4}x*\sqrt{1+x}dx\]

- anonymous

ok did you draw the region?

- anonymous

wait nevermind i made a mistake that simplified integral is wrong

- anonymous

but the first one is right

- anonymous

amistre64 sent me the pic http://assets.openstudy.com/updates/attachments/4de1249dc9a28b0b38e067c2-amistre64-1306602087960-likethis.jpg

- anonymous

the picture is a little strange the curve should be y=(x^2)/8, but it doesnt make much difference

- anonymous

so we know at the top limit:
x=sqrt(8y)
x^2=8y
r^2cos^2(theta)=8*r*sin(theta)
r=8sin(theta)/(cos^2(theta))
thats where the upper limit came from

- anonymous

and as a lower limit r=0

- anonymous

then the maximum r is 8sqrt(2) so we want to find corresponding theta
8sqrt(2)=8sin(theta)/cos^2(theta)
sqrt(2)=sin(theta)/cos^2(theta)
at Pi/4 sin(theta)=sqrt(2)/2 and cos^2(theta)=1/2
so Pi/4 is our solution.
theta from 0 to Pi/4

- anonymous

so you have that integral i sent you, and you need to solve that

- anonymous

the first integration is standard, but the second is a bit messy.

- anonymous

here is the correct final integral:
\[8^{3}/2\int\limits_{0}^{1}x \sqrt{1+x}dx\]

- anonymous

sorry should be 8^3/6

- anonymous

you can do the final integral by parts

- anonymous

if you want to know how I got to the integral above let me know

- anonymous

oh can u please tell me how u say that the maximum r is 8sqrt2

- anonymous

sure r is in the shaded region and the maximum is at the intersection of the two curves. there is a 45 45 90 right triangle with legs of size 8 so the hypotenuse is 8sqrt(2)

- anonymous

sqrt(8^2+8^2)=sqrt(128)=sqrt(64*2)=8sqrt(2)

- anonymous

does the rest of the initial set up make sense?

- anonymous

Need few minutes more..Im looking at your solution

- anonymous

when you do the first integration and plug in limits you get:
\[\int\limits_{0}^{\pi/4}8^{3}/3*(\sin^{3}(\theta)/\cos^{6}(\theta)d \theta\]

- anonymous

or:
\[8^{3}/3\int\limits_{0}^{\pi/4}\tan^{3}(\theta)\sec^{3}(\theta)d \theta\]

- anonymous

use substitution and chose tan^2(theta)=x
2tan(theta)sec^2(theta)=dx
1+tan^2(theta)=1+x
sec^2(theta)=1+x
sec(theta)=sqrt(1+x)
putting that all together and changing the limits of integration:
\[8^{3}/6\int\limits_{0}^{1}x \sqrt{1+x} dx\]

- anonymous

now by parts choose u=x dv=(1+x)^1/2
du=1 v=2((1+x)^(3/2))//3
and then you have a final simple power rule integral to solve

- anonymous

\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{8\sqrt{2}}r^2 drd \Theta\]

- anonymous

so what I have to do is solve this right

- anonymous

sorry i didnt mean the upper limit of integration on r was 8sqrt(2), I meant that was the r that let you find the limits of theta. the upper limit on r=8sin(theta)/cos^2(theta)

- anonymous

\[\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}dr\]
thats the inner integral

- anonymous

whole thing you need to solve:
\[\int\limits\limits_{0}^{\pi/4}\int\limits\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]

- anonymous

theta from 0 to Pi/4 you can see if you draw the picture correctly

- anonymous

oh i see.but like isnt solving the inner integral difficult?

- anonymous

no its not bad. Did you see my posts above, I wrote out the steps

- anonymous

let me check..I missed it i guess :) thanks a tons btw

- anonymous

you're welcome :) i want to be a math professor so this is fun for me.

- anonymous

btw hw dd u get 8^3/3 for r

- anonymous

thats from integrating r^2
int(r^2)=r^3/3. then plug in the upper and lower limits for r the constant term is 8^3/3

- anonymous

YEAH BUT THE upper and lower limits for r is 0 to sin(theeta)/cos^2(theta) right? that meas we have to substitue those two right?

- anonymous

0 to 8*sin(theta)/cos^2(theta).

- anonymous

there's an 8 in front of the second limit

- anonymous

so you get 8^3/3* sin^3(theta)/cos^6(theta)

- anonymous

then from there I posted most of the rest of the work

- anonymous

if you scroll up a bunch I wrote the work to find that upper limit of 8sin(theta)/cos^2(theta)

- anonymous

oh i geuss amistre's pic did make a difference he got the wrong bounds on theta

- anonymous

yeah I see..BTW if u dont mind me asking..are u a student or what? :)

- anonymous

Im a graduate student in math. Ill have a masters in a year and then probly go on to phd. How about you?

- anonymous

i have 3 years towards an engineering degree and a physics minor too, but I decided to do math

- anonymous

Engineering and Ohio State, math at Cleveland State. I am looking at University of Chicago, Michigan, Illinois, OSU, and some others for phd. Where are you studying?

- anonymous

EE is a lot of differential equations have you taken that yet?

- anonymous

engineering @ Ohio State I mean*

- anonymous

I'm Ryan. What nationality is your name? I've never seen it before

- anonymous

I've heard of Sri Lanka just not the name. Nice to meet you too:)
I'm looking through your school's math courses and it looks like you have a good math program there.

- anonymous

oh yeah..are college runs by quarter system and the pressure is a waaay tooo much to bear..:)

- anonymous

ohio state was quarters I didn't really like it

- anonymous

yeah...the speed is hard to bear..LOL..anyways I am new to this study room. Is this like fb I mean like can we add friends and stuff?

- anonymous

I've only been using it for a couple weeks I'm not really sure. I was trying to figure out if you could do that and have private chats, etc. but I don't think you can.

- anonymous

yeah it would have been great if they had that option..anyways I like to be friends with you..:) u have been really helpful for me..

- anonymous

the picture should come up. Smiley face drawn in paint. same as my icon here

- anonymous

k, sounds good. Ill talk to you later

- anonymous

sure thnx and see ya

- anonymous

hmm im on facebook now and it's not coming up

- anonymous

shud come up any second//

- anonymous

k

- anonymous

allright got it. cya!

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