Question

FInd volume of solid obtained by revolving around the y-axis the plane area between the graphy=1-x^2 and x-axis

Answer 1

define the interval

Answer 2

and since the revolutions produce circle with areas of pi r^2 ...

\[pi\int_{a}^{b}r^2\]r is defined by f(x), therefore
\[pi\int_{a}^{b}(f(x))^2~dx\]

Answer 3

n how will i find the intervals as there is no intervals given,is it the intersection point graph

Answer 4

the x axis is defined to be y=0

when does 1-x^2 = 0?

Answer 5

the x intercepts are defined to be where the graph crosses or touches the x axis ... the x intercepts of a graph are defined by y=0

Answer 6

and as the revolutions r around the y-axis so its interval would be c n d i think am i right that instead oof a n b on intergral sign it would be c n d

Answer 7

if we were to take this same setup and revolve around the y axis ... we would only use half the region, from 0 to 1 i believe

Answer 8

or shell it

Answer 9

ya graph depicts right

Answer 10

the area of the shells, are rectangles of of height f(x) and radius of x, so width of 2pix

Answer 11

so what would be the exact step by step solution for this volume problem

Answer 12

its a rough summary yes. i would leave it to you to hash out the specifics

Answer 13

thnx