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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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
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\[=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx\]
no thats not the region
the region is bounded by y = sqrt ( x + 4) , the y axis, its in quadrant 2

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

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