anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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dumbcow
  • dumbcow
\[=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx\]
anonymous
  • anonymous
no thats not the region
anonymous
  • anonymous
the region is bounded by y = sqrt ( x + 4) , the y axis, its in quadrant 2

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anonymous
  • anonymous
so using that integral , its from -4 to 0
dumbcow
  • dumbcow
yes but we are revolving around y-axis so it covers both quadrants symmetrically and this way you get a positive volume
anonymous
  • anonymous
it isnt symmertric though
dumbcow
  • dumbcow
x represents radius of 3-d figure, as you move away from y-axis the radius goes from 0 to 4
anonymous
  • anonymous
the area from 0 to 4 is not equal to the area from -4 to 0
anonymous
  • anonymous
how about this, revolve region bounded by y = sqrt x, x = 4, axis, about the line y = -4
dumbcow
  • dumbcow
oh i see, ok think of you reflected the function over y-axis replace x with (-x) f(x) = sqrt(4-x) i didn't do that in the intergral
anonymous
  • anonymous
yes, that will do it,
dumbcow
  • dumbcow
sorry
anonymous
  • anonymous
im trying to avoid these cases. are there other situations i can make up,
anonymous
  • anonymous
no you did fine :)
anonymous
  • anonymous
i dont seem to have a problem with disc method, since we have pi*r^2, but the shell method does produce negative
dumbcow
  • dumbcow
i guess always try to reflect onto positive quadrant when possible with volumes you want to integrate over positive values
dumbcow
  • dumbcow
make radius positive in shell method then adjust f(x) accordingly
anonymous
  • anonymous
but i think the disc method in that case works
anonymous
  • anonymous
using horizontal slices, we have x = y^2 - 4
anonymous
  • anonymous
for y = sqrt ( x + 4
dumbcow
  • dumbcow
yeah discmethod \[=\pi \int\limits_{0}^{2}(y^{2}-4)^{2}dy\]
anonymous
  • anonymous
so we get integral pi (
anonymous
  • anonymous
pi integral ( 0 - (y^2-4))^2
anonymous
  • anonymous
from -4 to 0
dumbcow
  • dumbcow
i prefer disc method
anonymous
  • anonymous
oh my bad
anonymous
  • anonymous
0 to 2

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