Question

find the are of a surface generated when the curve y=1/2x^3 + 1/2x^-1 from x=1 to x=3 is rotated about the x-axis.

Answer 1

Solid of revol problem.
Washer or shell?

Answer 2

is with surface formula
s=2pi int f(x) sqroot 91+(f(x))^2) dx

Answer 3

sorry no 9..
(1+(f(x))^2)

Answer 4

ohh thats shell ok cool.. i started but i am stuck...

Answer 5

This should be easy because we have no intersections to worry about
\[2\pi \int _a^bf(x) \sqrt{ f'(x)^2+1}dx\]

Answer 6

Find F'(x)

Answer 7

yea i have f'(x)=1/2x^2 - 1/2x^-2

Answer 8

\[F(x)=\frac{1}{2}x^2+\frac{1}{2}x^{-1}\]
wait is it the function?

Answer 9

the ftc is
1/6x^3 + 1/2x^-1

Answer 10

so
f'(x)=1/2x^2 - 1/2x^-2

Answer 11

square that

Answer 12

am i rigth?

Answer 13

ok
i got
1/4x^4 - 1/2 + 1/4x^-4

Answer 14

then i add 1

Answer 15

yes what do you have after adding one?

Answer 16

same different sign

Answer 17

(1/2x^2 + 1/2x^-2)^2

Answer 18

** different at the negative sign**

Answer 19

F'(x)^2=\[\left(-\frac{1}{2 x^2}+\frac{x^2}{2}\right)^2\]
1+f'(x)^2
\[1+\left(-\frac{1}{2 x^2}+\frac{x^2}{2}\right)^2\]
some algebra later
\[\frac{1+2 x^4+x^8}{4 x^4}\]

Answer 20

noo

Answer 21

wait

Answer 22

1/2x^2 + 1/2x ^-2

Answer 23

second term power negative

Answer 24

so is in the denominator

Answer 25

yes it is in denominator

Answer 26

u didnt do it like that y put it in the nominator'

Answer 27

\[F=\frac{1}{6}x^3+\frac{1}{2}x^{-1}\]
\[F'=\frac{1}{2}x^2-\frac{1}{2}x^{-2}\]
Rewritten as
\[F'=\frac{1}{2}x^2-\frac{1}{2x^2}\]

Answer 28

\[F'^2=\left(-\frac{1}{2 x^2}+\frac{x^2}{2}\right)^2\]
\[F'^2+1=1+\left(-\frac{1}{2 x^2}+\frac{x^2}{2}\right)^2\]
Rewritten as
\[\frac{1+2 x^4+x^8}{4 x^4}\]
which can be written as perfect square
\[\frac{\left(x^4+1\right)^2}{\left(2x^2\right)^2}\]
Since we are taking square root, we lose square

\[\frac{\left(x^4+1\right)}{\left(2x^2\right)}\]

Answer 29

\[2\pi \int _a^b\left(\frac{1}{6}x^3+\frac{1}{2}x^{-1}\right)\frac{\left(x^4+1\right)}{\left(2x^2\right)}dx\]
\[2\pi \int _a^b\frac{1}{4 x^3}+\frac{x}{3}+\frac{x^5}{12}dx\]
Simple integral from here

Answer 30

cool

Answer 31

i just finish the problem.. i had 209/9 pi sq units...lol