Question

**Calc2 Help** Surface Area generated by the curve y=x^3/2 from x = 0 to 3 about the x-axis.

Answer 1

so whats our integral look like?

Answer 2

\[2pi\int \sqrt{y^2(1+dy^2)}dx\]

Answer 3

der is 3y^2/3

Answer 4

i mean x..

Answer 5

3-2 = 1

3/2 x^1/2 is our derivative

Answer 6

http://www.wolframalpha.com/input/?i=derivative+x%5E3%2F2

Answer 7

\[2pi\int \sqrt{(x^{3/2})^2(1+(\frac{3}{2}x^{1/2})^2)}dx\]
\[2pi\int_{0}^{3} \sqrt{x^3(1+\frac{9}{4}x)}dx\]

Answer 8

you can square the 4, so now it's just pi outside the integral.

Answer 9

is that:\[x^{3/2}\]or \[\frac{1}{2}x^3\]???

Answer 10

where??

Answer 11

your y= x^3/2

Answer 12

is it:\[y=x^{3/2}\]
or
\[y=\frac12 x^3\]???

Answer 13

oh it is y = (x^3)/2

Answer 14

ahh, ok

Answer 15

let me fix up my int then :)
\[2pi\int \sqrt{(\frac{x^{3}}{2})^2(1+(\frac{3}{2}x^{2})^2)}dx\]
\[2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(1+(\frac{9}{4}x^{4}))}dx\]
\[2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(\frac{4+9x^4}{4})}dx\]
\[2pi\int_{0}^2 \sqrt{(\frac{4x^6+9x^{10}}{16})}dx\]
\[\frac{pi}{2}\int_{0}^2 \sqrt{4x^6+9x^{10}}dx\]

Answer 16

doesnt really pretty up does it :)

Answer 17

ok so how did you bring the x^3/2 inside the sqrt??

Answer 18

haha all these things are freaking ugly!!

Answer 19

i undid it, for example
\[\sqrt{a^2b}=a\sqrt{b}\]so to put it back in; square it

Answer 20

why did you not finish prettying it amistre?

Answer 21

I tried it and got an ugly big number..

Answer 22

lol, cause this math latex is slowing my computer like mole asses

Answer 23

lol\[\frac{\pi}2\int_{0}^{2}x^3\sqrt{4+9x^4}dx\]now it's beautiful!

Answer 24

\[2/3 * (4x^6 + 9x^(10))^(3/2)\]

Answer 25

still a bit yucky :) but im sure it trigs up nice

http://www.wolframalpha.com/input/?i=integrate+%28pi%2F2%29%28sqrt%284x%5E6%2B9x%5E10%29%29+from+0+to+2

Answer 26

sqrt(tan^2) if im not mistaken

Answer 27

@amistre.. 52.1416 didn't work. :/

Answer 28

im not the expert at plugging into programs .... :)

Answer 29

what trig?\[u=4+9x^4\to du=36x^3dx\]\[xdx=\frac{du}{72}\]\[\frac{\pi}{72}\int\sqrt udu\]

Answer 30

@Turing, that is what I had..

Answer 31

let me see what I get

Answer 32

I had pi/2 integral 2/3 * (-9x^4 + 1 )^ (3/2)

Answer 33

pi/72****

Answer 34

with limits 144 and..

Answer 35

1.. lol

Answer 36

yeah I got 52.14... as well
so I would have to read back and see what you guys were doing

Answer 37

I HATE THIS STUFF!!!!!!!

Answer 38

does yout program say to round it to some deci spot?

Answer 39

hey wait, I integrated from 0 to 2, not 3
let me try again...

Answer 40

It likes fractions..

Answer 41

577.041397813276 let me try to get a fraction...

Answer 42

it says 4 or 5 sig figs..

Answer 43

http://www.wolframalpha.com/input/?i=integrate+%282pi%29*%28x%5E3%2F2%29*%28sqrt%281%2B%283x%5E2%2F2%29%5E2%29%29+from+0+to+2

even without prettying it up and just plugging in the formula I get the same 52.

Answer 44

52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470

see if itll do that lol

Answer 45

but the integral is from 0 to 3, no?

Answer 46

52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470 HAHAHAHAHAAH.. THAT WORKED!!!

Answer 47

0 <= x <= 2

Answer 48

:)

Answer 49

Sorry for the caps, my excitement overtook my fingers.

Answer 50

well it is obviously right, but the post says 0 to 3...
I put in 0 to 2 at first as well, I must have picked up on it subconsciously

Answer 51

You guys are badass! Now it's time for me to study and to this crap on a test tomorrow morning.. Fun....

Answer 52

good luck :)

Answer 53

Thank ya! Take care guys.. See you next week with a new set of problems.. This time they will not be solids of revolution, but spring related force.