Question

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

Answer 1

Hoot! 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!

Answer 2

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

Answer 3

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

Answer 4

be right back .....

Answer 5

y <= x <= sqrt(8y)

0 <= y <= 8

Answer 6

{SS} sqrt(x^2 +y^2) dx.dy seems more appropriate than dr.dt

we gotta convert this to polar?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

This is the question though

Answer 11

nice :)

Answer 12

this is my picture

Answer 13

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

Answer 14

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

Answer 15

yeah I dont understand what my radius should be..

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

the slant of a rt tri with 45 degrees is sqrt(2); this goes 8 times longer...

so: 0 <= radius <= 8sqrt(2)

Answer 21

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

Answer 22

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

Answer 23

howdy sath :) im fine thnx

Answer 24

i wish i were there with you!

Answer 25

you just bored? or bad times?

Answer 26

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

Answer 27

atleast way to study.

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

x = r cos(t)
y = r sin(r) right

r cos(t) = sqrt(8y)

r cos(t) = sqrt(8 rsin(t))

Answer 33

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

Answer 34

or solve for r :)

Answer 35

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

Answer 36

this is sooo difficult.. :(

Answer 37

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

Answer 38

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

Answer 39

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)

Answer 40

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...

Answer 41

does that make sense?

Answer 42

so is the answer pi/16?

Answer 43

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 ...

Answer 44

yeah please...

Answer 45

hold on...

Answer 46

sure

Answer 47

well, i had to seek a higher source lol

Answer 48

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

Answer 49

but still it turns out to be incorrect..

Answer 50

is it looking for an exact? or an approximation?

Answer 51

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

Answer 52

i aint having no luck at it....

Answer 53

its fine then..thanks a lot :)

Answer 54

i have the answer exactly if you want.

1024/45+(1024/45)*sqrt(2)

Answer 55

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

Answer 56

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

Answer 57

can you read that?

Answer 58

ya i can read it..

Answer 59

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

Answer 60

i have it down to this:

\[8^3/6*\int\limits_{0}^{\pi/4}x*\sqrt{1+x}dx\]

Answer 61

ok did you draw the region?

Answer 62

wait nevermind i made a mistake that simplified integral is wrong

Answer 63

but the first one is right

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

and as a lower limit r=0

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

here is the correct final integral:

\[8^{3}/2\int\limits_{0}^{1}x \sqrt{1+x}dx\]

Answer 72

sorry should be 8^3/6

Answer 73

you can do the final integral by parts

Answer 74

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

Answer 75

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

Answer 76

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)

Answer 77

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

Answer 78

does the rest of the initial set up make sense?

Answer 79

Need few minutes more..Im looking at your solution

Answer 80

when you do the first integration and plug in limits you get:

\[\int\limits_{0}^{\pi/4}8^{3}/3*(\sin^{3}(\theta)/\cos^{6}(\theta)d \theta\]

Answer 81

or:

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

Answer 82

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:

\[8^{3}/6\int\limits_{0}^{1}x \sqrt{1+x} dx\]

Answer 83

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

Answer 84

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

Answer 85

so what I have to do is solve this right

Answer 86

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)

Answer 87

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

thats the inner integral

Answer 88

whole thing you need to solve:

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

Answer 89

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

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

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

Answer 94

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

Answer 95

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

Answer 96

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?