anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Owlfred
  • Owlfred
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amistre64
  • amistre64
double inting on polars eh .... sounds delightful :)
amistre64
  • amistre64
{SS} sqrt(x^2 +y^2) dr.dt

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amistre64
  • amistre64
be right back .....
amistre64
  • amistre64
y <= x <= sqrt(8y) 0 <= y <= 8
amistre64
  • amistre64
{SS} sqrt(x^2 +y^2) dx.dy seems more appropriate than dr.dt we gotta convert this to polar?
anonymous
  • anonymous
yeah we need to convert it to polar,,what i dnt understand is..How I should change my boundaries
amistre64
  • amistre64
id have to draw a picture..... :)
amistre64
  • amistre64
i know the cap is a sphere ... but for the domain I need to see it
anonymous
  • anonymous
This is the question though
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amistre64
  • amistre64
nice :)
amistre64
  • amistre64
this is my picture
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amistre64
  • 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
  • amistre64
and our radius goes from 0 to the longest distance there is along the y=x line; or 8sqrt(2)
anonymous
  • anonymous
yeah I dont understand what my radius should be..
amistre64
  • amistre64
if i see it right; your radius goes from 0 to 8sqrt(2)
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anonymous
  • anonymous
how do u say 8sqart2..i dnt understand that part
amistre64
  • amistre64
your radius is a constant change; your theta depends on sqrt(8y) right?
amistre64
  • 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
  • amistre64
the slant of a rt tri with 45 degrees is sqrt(2); this goes 8 times longer... so: 0 <= radius <= 8sqrt(2)
amistre64
  • amistre64
pi/4 <= theta <= sqrt(8y), just gotta figure out if that sqrt(8y) need to be altered...
anonymous
  • anonymous
hello anthony sir,,,how are you ??????you are excellent.
amistre64
  • amistre64
howdy sath :) im fine thnx
anonymous
  • anonymous
i wish i were there with you!
amistre64
  • amistre64
you just bored? or bad times?
anonymous
  • anonymous
nah...but am great of yours.wanna learn something from you.
anonymous
  • anonymous
atleast way to study.
anonymous
  • anonymous
im trying to calculate it and I got 256pi/3...but it turns out to be wrong :(
amistre64
  • amistre64
{SS} f(x,y) dA <=> {SS} f(rcos, rsin)r dr.dt
anonymous
  • 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
  • amistre64
i think the upper limit for the theta is spose to correspond to the x = sqrt(8y) function somehow
amistre64
  • amistre64
x = r cos(t) y = r sin(r) right r cos(t) = sqrt(8y) r cos(t) = sqrt(8 rsin(t))
amistre64
  • amistre64
r^2 cos^2(t) = 8r sin(t) solve for 't' i think
amistre64
  • amistre64
or solve for r :)
amistre64
  • amistre64
finding 't' sounds more reasonable to me; unless we can convert around to make it simpler
anonymous
  • anonymous
this is sooo difficult.. :(
amistre64
  • 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
  • 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
  • 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
  • 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
  • amistre64
does that make sense?
anonymous
  • anonymous
so is the answer pi/16?
amistre64
  • 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
  • anonymous
yeah please...
amistre64
  • amistre64
hold on...
anonymous
  • anonymous
sure
amistre64
  • amistre64
well, i had to seek a higher source lol
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amistre64
  • amistre64
http://www.wolframalpha.com/widgets/gallery/view.jsp?id=f5f3cbf14f4f5d6d2085bf2d0fb76e8a
anonymous
  • anonymous
but still it turns out to be incorrect..
amistre64
  • amistre64
is it looking for an exact? or an approximation?
anonymous
  • anonymous
I think we should keep the answers in terms of pi..
amistre64
  • amistre64
i aint having no luck at it....
anonymous
  • anonymous
its fine then..thanks a lot :)
anonymous
  • anonymous
i have the answer exactly if you want. 1024/45+(1024/45)*sqrt(2)
anonymous
  • 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
  • anonymous
\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]
anonymous
  • anonymous
can you read that?
anonymous
  • anonymous
ya i can read it..
anonymous
  • anonymous
omg its crct....pls tell me hw u do it
anonymous
  • anonymous
i have it down to this: \[8^3/6*\int\limits_{0}^{\pi/4}x*\sqrt{1+x}dx\]
anonymous
  • anonymous
ok did you draw the region?
anonymous
  • anonymous
wait nevermind i made a mistake that simplified integral is wrong
anonymous
  • anonymous
but the first one is right
anonymous
  • anonymous
amistre64 sent me the pic http://assets.openstudy.com/updates/attachments/4de1249dc9a28b0b38e067c2-amistre64-1306602087960-likethis.jpg
anonymous
  • anonymous
the picture is a little strange the curve should be y=(x^2)/8, but it doesnt make much difference
anonymous
  • 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
  • anonymous
and as a lower limit r=0
anonymous
  • 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
  • anonymous
so you have that integral i sent you, and you need to solve that
anonymous
  • anonymous
the first integration is standard, but the second is a bit messy.
anonymous
  • anonymous
here is the correct final integral: \[8^{3}/2\int\limits_{0}^{1}x \sqrt{1+x}dx\]
anonymous
  • anonymous
sorry should be 8^3/6
anonymous
  • anonymous
you can do the final integral by parts
anonymous
  • anonymous
if you want to know how I got to the integral above let me know
anonymous
  • anonymous
oh can u please tell me how u say that the maximum r is 8sqrt2
anonymous
  • 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
  • anonymous
sqrt(8^2+8^2)=sqrt(128)=sqrt(64*2)=8sqrt(2)
anonymous
  • anonymous
does the rest of the initial set up make sense?
anonymous
  • anonymous
Need few minutes more..Im looking at your solution
anonymous
  • 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
  • anonymous
or: \[8^{3}/3\int\limits_{0}^{\pi/4}\tan^{3}(\theta)\sec^{3}(\theta)d \theta\]
anonymous
  • 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
  • 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
  • anonymous
\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{8\sqrt{2}}r^2 drd \Theta\]
anonymous
  • anonymous
so what I have to do is solve this right
anonymous
  • 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
  • anonymous
\[\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}dr\] thats the inner integral
anonymous
  • 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
  • anonymous
theta from 0 to Pi/4 you can see if you draw the picture correctly
anonymous
  • anonymous
oh i see.but like isnt solving the inner integral difficult?
anonymous
  • anonymous
no its not bad. Did you see my posts above, I wrote out the steps
anonymous
  • anonymous
let me check..I missed it i guess :) thanks a tons btw
anonymous
  • anonymous
you're welcome :) i want to be a math professor so this is fun for me.
anonymous
  • anonymous
btw hw dd u get 8^3/3 for r
anonymous
  • 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
  • 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
  • anonymous
0 to 8*sin(theta)/cos^2(theta).
anonymous
  • anonymous
there's an 8 in front of the second limit
anonymous
  • anonymous
so you get 8^3/3* sin^3(theta)/cos^6(theta)
anonymous
  • anonymous
then from there I posted most of the rest of the work
anonymous
  • anonymous
if you scroll up a bunch I wrote the work to find that upper limit of 8sin(theta)/cos^2(theta)
anonymous
  • anonymous
oh i geuss amistre's pic did make a difference he got the wrong bounds on theta
anonymous
  • anonymous
yeah I see..BTW if u dont mind me asking..are u a student or what? :)
anonymous
  • 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
  • anonymous
i have 3 years towards an engineering degree and a physics minor too, but I decided to do math
anonymous
  • 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
  • anonymous
EE is a lot of differential equations have you taken that yet?
anonymous
  • anonymous
engineering @ Ohio State I mean*
anonymous
  • anonymous
I'm Ryan. What nationality is your name? I've never seen it before
anonymous
  • 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
  • anonymous
oh yeah..are college runs by quarter system and the pressure is a waaay tooo much to bear..:)
anonymous
  • anonymous
ohio state was quarters I didn't really like it
anonymous
  • 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
  • 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
  • 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
  • anonymous
the picture should come up. Smiley face drawn in paint. same as my icon here
anonymous
  • anonymous
k, sounds good. Ill talk to you later
anonymous
  • anonymous
sure thnx and see ya
anonymous
  • anonymous
hmm im on facebook now and it's not coming up
anonymous
  • anonymous
shud come up any second//
anonymous
  • anonymous
k
anonymous
  • anonymous
allright got it. cya!

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