Question

let r be the region in the first quadrant bounded by the graph y=8- x^ (3/2) Find the area of the region R . Find the volume of the solid generated when R is revolved about the x-axis

Answer 1

you sure its x^(3/2) ?

Answer 2

the function itself is of no consequence; it only helps define the radial limit and the boundary for integration

Answer 3

when does y = 0?

Answer 4

yes its x^3/2

Answer 5

i typed allthe given info

Answer 6

ok, cause the last question i just did on this was an x^2 there :)

Answer 7

yes, and all the given info allows us to find the needed info

Answer 8

when does y=0?

Answer 9

i'm so lost

Answer 10

\[\int_{y=8}^{y=0}(area.of.a.circle)dx\]

Answer 11

how did you get that?

Answer 12

y = 8-x^(3/2); when does y=0?

0 = 8-x^(3/2) ; solve for x

Answer 13

thats a guide; thats what we are going to have to fill in to answer the question

Answer 14

x=2?

Answer 15

o wait no

Answer 16

well, if were gonna be guessing; lets at least fill it in to see :)

Answer 17

x=4?

Answer 18

4 is better

Answer 19

i forgot to square it, my bad

Answer 20

\[\int_{0}^{4}(area.function)dx\]

Answer 21

each rotation produces a circle; what the formula for the area of a circle?

Answer 22

pi r sqd

Answer 23

good; except for in this case; r = f(x) as we move down the line

Answer 24

so how do we get r

Answer 25

\[\int_{0}^{4}\pi[r]^2dx\]
\[\int_{0}^{4}\pi[f(x)]^2dx\]

Answer 26

f(x)=y

Answer 27

subsitute 4 for x?