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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0double inting on polars eh .... sounds delightful :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{SS} sqrt(x^2 +y^2) dr.dt

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y <= x <= sqrt(8y) 0 <= y <= 8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{SS} sqrt(x^2 +y^2) dx.dy seems more appropriate than dr.dt we gotta convert this to polar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah we need to convert it to polar,,what i dnt understand is..How I should change my boundaries

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id have to draw a picture..... :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i know the cap is a sphere ... but for the domain I need to see it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the question though

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as we sweep this across our dt doesnt really change does it?it goes for pi/4 to pi/2 if anything right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and our radius goes from 0 to the longest distance there is along the y=x line; or 8sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah I dont understand what my radius should be..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if i see it right; your radius goes from 0 to 8sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do u say 8sqart2..i dnt understand that part

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your radius is a constant change; your theta depends on sqrt(8y) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok... 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
 5 years ago
Best ResponseYou've already chosen the best response.0the slant of a rt tri with 45 degrees is sqrt(2); this goes 8 times longer... so: 0 <= radius <= 8sqrt(2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pi/4 <= theta <= sqrt(8y), just gotta figure out if that sqrt(8y) need to be altered...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello anthony sir,,,how are you ??????you are excellent.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0howdy sath :) im fine thnx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wish i were there with you!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you just bored? or bad times?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nah...but am great of yours.wanna learn something from you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0atleast way to study.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to calculate it and I got 256pi/3...but it turns out to be wrong :(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{SS} f(x,y) dA <=> {SS} f(rcos, rsin)r dr.dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i have to integrate this right int(int(r^2, r = 0 .. 8*sqrt(2)), theta = (1/4)*Pi .. (1/2)*Pi)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i think the upper limit for the theta is spose to correspond to the x = sqrt(8y) function somehow

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = r cos(t) y = r sin(r) right r cos(t) = sqrt(8y) r cos(t) = sqrt(8 rsin(t))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0r^2 cos^2(t) = 8r sin(t) solve for 't' i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0finding 't' sounds more reasonable to me; unless we can convert around to make it simpler

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is sooo difficult.. :(

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah..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
 5 years ago
Best ResponseYou've already chosen the best response.0r^2 cos^2(t) = 8r sin(t) ; /r r cos^2(t) = 8 sin(t) ; indentity r (1sin^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
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is the answer pi/16?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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 ...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, i had to seek a higher source lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/widgets/gallery/view.jsp?id=f5f3cbf14f4f5d6d2085bf2d0fb76e8a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but still it turns out to be incorrect..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is it looking for an exact? or an approximation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think we should keep the answers in terms of pi..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i aint having no luck at it....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its fine then..thanks a lot :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have the answer exactly if you want. 1024/45+(1024/45)*sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i have the correct integral. I used maple to solve though. I can try to do it by hand if you want.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg its crct....pls tell me hw u do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have it down to this: \[8^3/6*\int\limits_{0}^{\pi/4}x*\sqrt{1+x}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok did you draw the region?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait nevermind i made a mistake that simplified integral is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the first one is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre64 sent me the pic http://assets.openstudy.com/updates/attachments/4de1249dc9a28b0b38e067c2amistre641306602087960likethis.jpg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the picture is a little strange the curve should be y=(x^2)/8, but it doesnt make much difference

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0and as a lower limit r=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.0so you have that integral i sent you, and you need to solve that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first integration is standard, but the second is a bit messy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is the correct final integral: \[8^{3}/2\int\limits_{0}^{1}x \sqrt{1+x}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry should be 8^3/6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do the final integral by parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you want to know how I got to the integral above let me know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh can u please tell me how u say that the maximum r is 8sqrt2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure 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
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(8^2+8^2)=sqrt(128)=sqrt(64*2)=8sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does the rest of the initial set up make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Need few minutes more..Im looking at your solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0or: \[8^{3}/3\int\limits_{0}^{\pi/4}\tan^{3}(\theta)\sec^{3}(\theta)d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use 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
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{8\sqrt{2}}r^2 drd \Theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what I have to do is solve this right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}dr\] thats the inner integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whole 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
 5 years ago
Best ResponseYou've already chosen the best response.0theta from 0 to Pi/4 you can see if you draw the picture correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see.but like isnt solving the inner integral difficult?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no its not bad. Did you see my posts above, I wrote out the steps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me check..I missed it i guess :) thanks a tons btw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're welcome :) i want to be a math professor so this is fun for me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw hw dd u get 8^3/3 for r

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats 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
 5 years ago
Best ResponseYou've already chosen the best response.0YEAH 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
 5 years ago
Best ResponseYou've already chosen the best response.00 to 8*sin(theta)/cos^2(theta).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there's an 8 in front of the second limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you get 8^3/3* sin^3(theta)/cos^6(theta)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then from there I posted most of the rest of the work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you scroll up a bunch I wrote the work to find that upper limit of 8sin(theta)/cos^2(theta)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i geuss amistre's pic did make a difference he got the wrong bounds on theta

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah I see..BTW if u dont mind me asking..are u a student or what? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im a graduate student in math. Ill have a masters in a year and then probly go on to phd. How about you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have 3 years towards an engineering degree and a physics minor too, but I decided to do math

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Engineering 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
 5 years ago
Best ResponseYou've already chosen the best response.0EE is a lot of differential equations have you taken that yet?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0engineering @ Ohio State I mean*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm Ryan. What nationality is your name? I've never seen it before

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah..are college runs by quarter system and the pressure is a waaay tooo much to bear..:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohio state was quarters I didn't really like it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah...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
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 5 years ago
Best ResponseYou've already chosen the best response.0the picture should come up. Smiley face drawn in paint. same as my icon here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k, sounds good. Ill talk to you later

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm im on facebook now and it's not coming up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shud come up any second//

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0allright got it. cya!
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