anonymous
  • anonymous
**Calc2 Help** Surface Area generated by the curve y=x^3/2 from x = 0 to 3 about the x-axis.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
so whats our integral look like?
amistre64
  • amistre64
\[2pi\int \sqrt{y^2(1+dy^2)}dx\]
anonymous
  • anonymous
der is 3y^2/3

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anonymous
  • anonymous
i mean x..
amistre64
  • amistre64
3-2 = 1 3/2 x^1/2 is our derivative
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=derivative+x%5E3%2F2
amistre64
  • amistre64
\[2pi\int \sqrt{(x^{3/2})^2(1+(\frac{3}{2}x^{1/2})^2)}dx\] \[2pi\int_{0}^{3} \sqrt{x^3(1+\frac{9}{4}x)}dx\]
anonymous
  • anonymous
you can square the 4, so now it's just pi outside the integral.
amistre64
  • amistre64
is that:\[x^{3/2}\]or \[\frac{1}{2}x^3\]???
anonymous
  • anonymous
where??
amistre64
  • amistre64
your y= x^3/2
amistre64
  • amistre64
is it:\[y=x^{3/2}\] or \[y=\frac12 x^3\]???
anonymous
  • anonymous
oh it is y = (x^3)/2
amistre64
  • amistre64
ahh, ok
anonymous
  • anonymous
1 Attachment
amistre64
  • amistre64
let me fix up my int then :) \[2pi\int \sqrt{(\frac{x^{3}}{2})^2(1+(\frac{3}{2}x^{2})^2)}dx\] \[2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(1+(\frac{9}{4}x^{4}))}dx\] \[2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(\frac{4+9x^4}{4})}dx\] \[2pi\int_{0}^2 \sqrt{(\frac{4x^6+9x^{10}}{16})}dx\] \[\frac{pi}{2}\int_{0}^2 \sqrt{4x^6+9x^{10}}dx\]
amistre64
  • amistre64
doesnt really pretty up does it :)
anonymous
  • anonymous
ok so how did you bring the x^3/2 inside the sqrt??
anonymous
  • anonymous
haha all these things are freaking ugly!!
amistre64
  • amistre64
i undid it, for example \[\sqrt{a^2b}=a\sqrt{b}\]so to put it back in; square it
TuringTest
  • TuringTest
why did you not finish prettying it amistre?
anonymous
  • anonymous
I tried it and got an ugly big number..
amistre64
  • amistre64
lol, cause this math latex is slowing my computer like mole asses
TuringTest
  • TuringTest
lol\[\frac{\pi}2\int_{0}^{2}x^3\sqrt{4+9x^4}dx\]now it's beautiful!
anonymous
  • anonymous
\[2/3 * (4x^6 + 9x^(10))^(3/2)\]
amistre64
  • amistre64
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
amistre64
  • amistre64
sqrt(tan^2) if im not mistaken
anonymous
  • anonymous
@amistre.. 52.1416 didn't work. :/
amistre64
  • amistre64
im not the expert at plugging into programs .... :)
TuringTest
  • TuringTest
what trig?\[u=4+9x^4\to du=36x^3dx\]\[xdx=\frac{du}{72}\]\[\frac{\pi}{72}\int\sqrt udu\]
anonymous
  • anonymous
@Turing, that is what I had..
TuringTest
  • TuringTest
let me see what I get
anonymous
  • anonymous
I had pi/2 integral 2/3 * (-9x^4 + 1 )^ (3/2)
anonymous
  • anonymous
pi/72****
anonymous
  • anonymous
with limits 144 and..
anonymous
  • anonymous
1.. lol
TuringTest
  • TuringTest
yeah I got 52.14... as well so I would have to read back and see what you guys were doing
anonymous
  • anonymous
I HATE THIS STUFF!!!!!!!
amistre64
  • amistre64
does yout program say to round it to some deci spot?
TuringTest
  • TuringTest
hey wait, I integrated from 0 to 2, not 3 let me try again...
anonymous
  • anonymous
It likes fractions..
TuringTest
  • TuringTest
577.041397813276 let me try to get a fraction...
anonymous
  • anonymous
it says 4 or 5 sig figs..
amistre64
  • amistre64
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.
amistre64
  • amistre64
52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470 see if itll do that lol
TuringTest
  • TuringTest
but the integral is from 0 to 3, no?
anonymous
  • anonymous
52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470 HAHAHAHAHAAH.. THAT WORKED!!!
amistre64
  • amistre64
0 <= x <= 2
amistre64
  • amistre64
:)
anonymous
  • anonymous
Sorry for the caps, my excitement overtook my fingers.
TuringTest
  • TuringTest
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
anonymous
  • anonymous
You guys are badass! Now it's time for me to study and to this crap on a test tomorrow morning.. Fun....
amistre64
  • amistre64
good luck :)
anonymous
  • anonymous
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.

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