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anonymous
 5 years ago
revolve the region bounded by x axis , y = sqrt ( x+ 4) , and x=0 about y axis using shell method.
compare it to the disc method
anonymous
 5 years ago
revolve the region bounded by x axis , y = sqrt ( x+ 4) , and x=0 about y axis using shell method. compare it to the disc method

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no thats not the region

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the region is bounded by y = sqrt ( x + 4) , the y axis, its in quadrant 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so using that integral , its from 4 to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but we are revolving around yaxis so it covers both quadrants symmetrically and this way you get a positive volume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it isnt symmertric though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x represents radius of 3d figure, as you move away from yaxis the radius goes from 0 to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the area from 0 to 4 is not equal to the area from 4 to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about this, revolve region bounded by y = sqrt x, x = 4, axis, about the line y = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see, ok think of you reflected the function over yaxis replace x with (x) f(x) = sqrt(4x) i didn't do that in the intergral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that will do it,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to avoid these cases. are there other situations i can make up,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont seem to have a problem with disc method, since we have pi*r^2, but the shell method does produce negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess always try to reflect onto positive quadrant when possible with volumes you want to integrate over positive values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0make radius positive in shell method then adjust f(x) accordingly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i think the disc method in that case works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using horizontal slices, we have x = y^2  4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah discmethod \[=\pi \int\limits_{0}^{2}(y^{2}4)^{2}dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we get integral pi (

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pi integral ( 0  (y^24))^2
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