Question

Find the area of the surface obtained by rotating the curve y= \sqrt{3 x} from x=0 to x = 3 about the x-axis.

Answer 1

first draw the curves

Answer 2

Doin'

Answer 3

How precise do I need to be?

Answer 4

I have 0 and 3 when I draw the sqrt function.

Answer 5

crude
just so you see what you are doing :)

Answer 6

Got it.

Answer 7

then
\[
S=\int_a^b2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
\]

Answer 8

Is dy/dx the \[\sqrt{3x}\]

Answer 9

?

Answer 10

Or its derivative?

Answer 11

no. y=sqrt(3x)
you've gotta find dy/dx

Answer 12

-3/(2*sqrt(3x))

Answer 13

Okay, cool

Answer 14

So, I get this gargantuan thing...

Answer 15

\[\int\limits_{0}^{3} 2\pi \sqrt{3x}\sqrt{1+1/12x}\]

Answer 16

Do I just plug in 3 and 0?

Answer 17

the second radial has a major error

Answer 18

Radial?

Answer 19

What's a radial?

Answer 20

\[\sqrt{1+\left(\frac{-3}{2\sqrt{3x}}\right)^2}=\sqrt{1+{3\over2x}}=\sqrt{\frac{2x+3}{2x}}
\]
**radical**

Answer 21

\[\sqrt{3x}\sqrt{\frac{2x+3}{2x}}=\sqrt{{3\over2}(x+3)}
\]

Answer 22

\[
S=2\pi\sqrt{3\over2}\int_0^3\sqrt{x+3}dx
\]
and solve

Answer 23

Where is the negative 3 coming from?

Answer 24

oh yes.. my bad.. no -ve
but its squared. so I got lucked out ;)

Answer 25

What?

Answer 26

you are talking about the "-3" in the radical in the first step, right?

Answer 27

Yes.

Answer 28

yes and it is squared.. so (-3)^2=+9 so, just ignore that -ve sign (take it as +ve like you said)

Answer 29

Why is a 3 there though? Even if it's positive? 'Cause the derivative of sqrt3x is 1/2sqrt3x.

Answer 30

chain rule, pal

Answer 31

Chain rule? Ooooooooohhhhh....

Answer 32

I'm studying for a calc final (252) and my prof. handed us 38 problems to do in two days. The little things are slipping me.

Answer 33

A lot.

Answer 34

gotta be careful though

Answer 35

Oh yes.

Answer 36

So, I get it until.....

Answer 37

\[\sqrt{3x}\sqrt{2x+3/2x}\]

Answer 38

I got something totally different. I got


\[\sqrt{3x+9x/2x}\]

Answer 39

ok that is fine too. but take the LCM of the fraction

Answer 40

\[\large 2\pi \int\limits_0^3 \sqrt{3x}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx \qquad = \qquad 2\pi \int\limits_0^3 \sqrt{3x}\sqrt{1+\left(\frac{3}{2\sqrt{3x}}\right)^2}dx\]

\[\large 2\pi \int\limits_0^3 \sqrt{3x}\sqrt{1+\left(\frac{9}{12x}\right)}dx \qquad = \qquad 2\pi \int\limits_0^3 \sqrt{3x}\sqrt{\frac{12x+9}{12x}}dx\]

\[\large 2\pi \int\limits\limits_0^3 \sqrt{3x}\frac{\sqrt{12x+9}}{2\sqrt{3x}}dx\]

Answer 41

Hmm I'm trying to figure out how you got to the spot you're at.

Answer 42

i mis-wrote it.. instead of 12, i put a "2"

Answer 43

Yeah I forgot the 12 too. Tell ya what...you guys reply and I'll check back later. My brain is burning out I think. I'll be back.

Answer 44

\[\large \cancel2\pi \int\limits\limits\limits_0^3 \cancel{\sqrt{3x}}\frac{\sqrt{12x+9}}{\cancel2\cancel{\sqrt{3x}}}dx\qquad = \qquad \pi \int\limits_0^3 \sqrt{12x+9}dx\]

Answer 45

I think you end up with something like this.

Answer 46

haha. hmm .. its late

Answer 47

\[
S=2\pi\int_0^3\sqrt{3x\frac{12x+9}{12x}}dx\\
=2\pi\int_0^3\frac{\sqrt{12x+9}}{2}dx\\
12x+9=u\quad,dx=du/12\quad,9\le u\le45\\
S=\pi\int_9^{45}u^{1/2}{du\over12}={\pi\over12}{2\over3}\left[u\sqrt{u}\right]_9^{45}\\
={\pi\over18}[45\sqrt{45}-9\cdot3]\\
S={\pi\over2}(5\sqrt{45}-3)
\]