## 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

1. Owlfred

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!

2. amistre64

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

3. amistre64

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

4. amistre64

be right back .....

5. amistre64

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

6. amistre64

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

7. anonymous

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

8. amistre64

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

9. amistre64

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

10. anonymous

This is the question though

11. amistre64

nice :)

12. amistre64

this is my picture

13. amistre64

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

14. amistre64

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

15. anonymous

yeah I dont understand what my radius should be..

16. amistre64

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

17. anonymous

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

18. amistre64

19. 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

20. amistre64

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

21. amistre64

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

22. anonymous

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

23. amistre64

howdy sath :) im fine thnx

24. anonymous

i wish i were there with you!

25. amistre64

you just bored? or bad times?

26. anonymous

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

27. anonymous

atleast way to study.

28. anonymous

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

29. amistre64

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

30. anonymous

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

31. amistre64

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

32. amistre64

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

33. amistre64

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

34. amistre64

or solve for r :)

35. amistre64

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

36. anonymous

this is sooo difficult.. :(

37. 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

38. 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

39. 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)

40. 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...

41. amistre64

does that make sense?

42. anonymous

43. 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 ...

44. anonymous

45. amistre64

hold on...

46. anonymous

sure

47. amistre64

well, i had to seek a higher source lol

48. amistre64
49. anonymous

but still it turns out to be incorrect..

50. amistre64

is it looking for an exact? or an approximation?

51. anonymous

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

52. amistre64

i aint having no luck at it....

53. anonymous

its fine then..thanks a lot :)

54. anonymous

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

55. anonymous

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

56. anonymous

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

57. anonymous

58. anonymous

59. anonymous

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

60. anonymous

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

61. anonymous

ok did you draw the region?

62. anonymous

wait nevermind i made a mistake that simplified integral is wrong

63. anonymous

but the first one is right

64. anonymous
65. anonymous

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

66. 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

67. anonymous

and as a lower limit r=0

68. 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

69. anonymous

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

70. anonymous

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

71. anonymous

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

72. anonymous

sorry should be 8^3/6

73. anonymous

you can do the final integral by parts

74. anonymous

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

75. anonymous

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

76. 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)

77. anonymous

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

78. anonymous

does the rest of the initial set up make sense?

79. anonymous

Need few minutes more..Im looking at your solution

80. anonymous

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

81. anonymous

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

82. 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: $8^{3}/6\int\limits_{0}^{1}x \sqrt{1+x} dx$

83. 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

84. anonymous

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

85. anonymous

so what I have to do is solve this right

86. 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)

87. anonymous

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

88. anonymous

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

89. anonymous

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

90. anonymous

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

91. anonymous

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

92. anonymous

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

93. anonymous

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

94. anonymous

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

95. 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

96. 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?

97. anonymous

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

98. anonymous

there's an 8 in front of the second limit

99. anonymous

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

100. anonymous

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

101. anonymous

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

102. anonymous

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

103. anonymous

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

104. anonymous

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

105. anonymous

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

106. 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?

107. anonymous

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

108. anonymous

engineering @ Ohio State I mean*

109. anonymous

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

110. 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.

111. anonymous

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

112. anonymous

ohio state was quarters I didn't really like it

113. 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?

114. 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.

115. 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..

116. anonymous

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

117. anonymous

k, sounds good. Ill talk to you later

118. anonymous

sure thnx and see ya

119. anonymous

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

120. anonymous

shud come up any second//

121. anonymous

k

122. anonymous

allright got it. cya!