Question

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

Answer 1

\[=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx\]

Answer 2

no thats not the region

Answer 3

the region is bounded by y = sqrt ( x + 4) , the y axis,
its in quadrant 2

Answer 4

so using that integral , its from -4 to 0

Answer 5

yes but we are revolving around y-axis so it covers both quadrants symmetrically
and this way you get a positive volume

Answer 6

it isnt symmertric though

Answer 7

x represents radius of 3-d figure, as you move away from y-axis the radius goes from 0 to 4

Answer 8

the area from 0 to 4 is not equal to the area from -4 to 0

Answer 9

how about this, revolve region bounded by y = sqrt x, x = 4, axis, about the line y = -4

Answer 10

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

Answer 11

yes, that will do it,

Answer 12

sorry

Answer 13

im trying to avoid these cases. are there other situations i can make up,

Answer 14

no you did fine :)

Answer 15

i dont seem to have a problem with disc method, since we have pi*r^2, but the shell method does produce negative

Answer 16

i guess always try to reflect onto positive quadrant when possible
with volumes you want to integrate over positive values

Answer 17

make radius positive in shell method then adjust f(x) accordingly

Answer 18

but i think the disc method in that case works

Answer 19

using horizontal slices, we have x = y^2 - 4

Answer 20

for y = sqrt ( x + 4

Answer 21

yeah discmethod
\[=\pi \int\limits_{0}^{2}(y^{2}-4)^{2}dy\]

Answer 22

so we get integral pi (

Answer 23

pi integral ( 0 - (y^2-4))^2

Answer 24

from -4 to 0

Answer 25

i prefer disc method

Answer 26

oh my bad

Answer 27

0 to 2