Question

whats the volume of the solid obtained when rotating the region under the graph of the function f(x)= x^(13/12)?? on the interval [1,3]??? need help

Answer 1

vol = INT pi f(x)^2 dx between x = 1 and 3

= INT pi x^169/144 dx
= pi [x^313/144 * 144/313] btw 1 and 3
=
= pi [3^313/144 * 144/313 -1^313/144 * 144/313 ]

Answer 2

OK - can u follow it ?? its an awkward one as u have to square
x^(13/12)

Answer 3

ya see i already had all that but its telling me i got the wrong answer

Answer 4

the vol of revolution = INT pi f(x)^2 dx

Answer 5

did u make a mistake in the calculation perhaps what was the value the book gave?

Answer 6

(24.573pi/13

Answer 7

ok i'll work out what i've got

Answer 8

ya thats not even the right answer either

Answer 9

well i'm pretty sure the integration is correct but i've got to go - sorry creeg - hope u get good answer - i'll check back in an hour to see if u still want help

Answer 10

ill be here

Answer 11

ok - i'm back and i can see i messed up on squaring x^ 13/12 its not x^169/144 you just multiply the power by 2 so

x ^(13/12) squared = x ^ (26/12) = x ^ (13/6)
so the working is:-

INT pi x^13/6 dx
= pi [x^19/6 * 6/19] btw 1 and 3
=
= pi [3^19/6 * 6/19 -1^19/6 * 6/19]

using my calculator I get

=9.924 pi