Question

Find the volume of the solid obtained by rotating the region enclosed between y=x^(1/2)+1, the line y=2 and the y-axis about the y-axis.

Thanks.

Answer 1

I can take a try at this but I'm rusty so bear with me

Answer 2

cylindrical shell?

Answer 3

yeah that is what I would use

Answer 4

oh ok well it's all yours @Australopithecus :)

Answer 5

isnt cylindrical disk useable?

Answer 6

Yeah I'm pretty sure you can do both methods

Answer 7

set
2 = x^(1/2)+1

solve for x

(1)^2 = x
x = 1
so we have a point of intersection at (1,2) for both functions, and the other intersection point is (0,1)



Thus, we integrate from 1 to 2 in terms of y

write the function in terms of y
y = x^(1/2)+1
(y+1)^2 = x

then the formula for cylindrical shells is 2(pi)r(h)
h = (y+1)^2
r = 2 - (y+1)^2 - 1
r = 1 - (y+1)^2


Thus it would be,

integral from 1 to 2 of 2pi(1 - (y+1)^2)((y+1)^2)dy

I'm probably wrong

@alexwee123

Answer 8

I treated it as a solid of revolution,
A(y) = pi((y-1)^2)^2
= pi(y-1)^4
V = integral 2 to 1 of A(y) = pi/5

But I know if I am right?

Answer 9

@KingGeorge

can you please set us straight on this question

Answer 10

@t1ger5, you are correct :)
if you use disc method here you must integrate with respect to y
if you use shell method , integrate with respect to x
@Australopithecus
you had the right idea except you put it in terms of the wrong variable