anonymous
  • anonymous
Evaluate ∫ ∫ F.ndS if F=yi-xj+8k and the surface S is that part of the sphere x^2 + y^2 + z^2=9. That is above the region d in the xy-plane enclosed by the circle x^2 + y^2=4
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dan815
  • dan815
hi
dan815
  • dan815
|dw:1377702543204:dw|
dan815
  • dan815
|dw:1377702784235:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dan815
  • dan815
|dw:1377702827399:dw|
dan815
  • dan815
|dw:1377702992878:dw|
dan815
  • dan815
|dw:1377703251262:dw|
dan815
  • dan815
|dw:1377706278157:dw|
dan815
  • dan815
|dw:1377706353108:dw|
dan815
  • dan815
rho is a constant so
dan815
  • dan815
|dw:1377706651069:dw|
dan815
  • dan815
@FutureMathProfessor check this answer over!! feel a bit rusty with surface integrals
dan815
  • dan815
@DebbieG can you check my answer over plz
ybarrap
  • ybarrap
$$ \text{The Surface: }f(x,y,z)=x^2+y+2+z^2=9 \\ \text{The Field: }\bar{F}=y\hat{i}-x\hat{j}+8\hat{k}\\ \text{The Normal Vector: }\hat{n}={\triangle f \over |\triangle f| }\\={2x\hat{i}+2y\hat{j}+2z\hat{k}\over \sqrt{4x^2+4y^2+4z^2}}=2{x\hat{i}+y\hat{j}+z\hat{k}\over 2\sqrt{x^2+y^2+z^2}}={x\hat{i}+y\hat{j}+z\hat{k}\over \sqrt{9}}\\={x\hat{i}+y\hat{j}+z\hat{k}\over 3}\\ \hat{F}\cdot\hat{n= }(y\hat{i}-x\hat{j}+8\hat{k})\cdot {x\hat{i}+y\hat{j}+z\hat{k}\over 3}\\ =\frac{8z}{3}\\ \text{z= }\sqrt{9-x^2-y^2}\\ \text{Let r= }x^2+y^2\\ \text{then z =}\sqrt{9-r^2}\\ \text{So, }\iint_{x^2+y^2\le 4}\hat{F}\cdot\hat{n}~dA=\int_{0\le\theta\le 2\pi}\int_{r\le 2}\frac 8 3 \sqrt{9-r^2}~r~dr~d\theta\\ \frac 8 3 \int_{0\le\theta\le 2\pi}\left (9 - \frac{5\sqrt 5}{3} \right )d\theta\\ =\frac 8 3 \left (9 - \frac{5\sqrt 5}{3} \right )2\pi\\ =16\pi \left (3 - \frac{5\sqrt 5}{9} \right )\approx 128 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\blacksquare } $$
dan815
  • dan815
@ybarrap are you sure u can do that?? arebt you just integrating over a circle, area when we had this kind of |dw:1377734732000:dw|
dan815
  • dan815
whats my mistake if the answer isnt supposed to be 64 pi
dan815
  • dan815
oh umm also your answer evaluates to 88. something
dan815
  • dan815
but anyway can you explain a bit of your thought process
dan815
  • dan815
u canculated the f.n thats fine, then ur trying to this in cylindrical coordinates
dan815
  • dan815
in which case wont you have to incorporate z axis for your surface area element wont you
dan815
  • dan815
unless you do cross and find DA element interms of DS and its differential part
dan815
  • dan815
|dw:1377735392978:dw|
dan815
  • dan815
youd have to do that conversion to da element wont youj
dan815
  • dan815
|dw:1377735537747:dw|
dan815
  • dan815
in this case youd get a simpler equation not converting that n into unit normal
dan815
  • dan815
|dw:1377735614832:dw|
dan815
  • dan815
same direction yes but not sure if those magnitudes would be the same
dan815
  • dan815
then you would get
dan815
  • dan815
basically whatever you got multiplied by 3 xD but still not sure if you can do that cancellation i did with the normal magnitude like i did in there
dan815
  • dan815
im getting 201 tho or 64 pi leme see if u can really make that cancellation
dan815
  • dan815
|dw:1377736395809:dw|
dan815
  • dan815
|dw:1377736615440:dw|
dan815
  • dan815
|dw:1377736700673:dw|
dan815
  • dan815
|dw:1377736722562:dw|
dan815
  • dan815
|dw:1377736961317:dw|
dan815
  • dan815
o snap this simplies to this again
dan815
  • dan815
this must be the right answe!! i wanna re do the spherical coordinate then!! why am i getting only 201
dan815
  • dan815
|dw:1377737705243:dw|
dan815
  • dan815
OH MY GOODNES 32pi now lol
dan815
  • dan815
this is getting bad
dan815
  • dan815
|dw:1377738076808:dw|