Question

Plz Help!

Answer 1

with what?

Answer 2

I'm writing, just wait

Answer 3

\[dx/dt=t \sqrt{t ^{2}+4}, dy/dt=-t \sqrt{4-t ^{2}}\] FInd the surface ara of revolution when the curve described above is completely rotated from t=0 to t=2, around the x-axis

Answer 4

@amistre64 I know you are a powerful mathematician, so please give me a hand

Answer 5

parametrics eh ....

Answer 6

i recall this might have something to with the ds:\[ds=\sqrt{(x')^2+(y')^2}\]

Answer 7

\[\int2\pi y~ds\]

Answer 8

Actually, I 've gone as far as writing it as S=\[\int\limits_{0}^{2} t(4-t ^{2})^{\frac{ 3 }{ 2}} dt \] but i don't know how to proceed

Answer 9

id like to see your work leading to that point :)

Answer 10

sorry, but it would take me a million years to write the whole process in this web

Answer 11

I'm pretty sure my result is correct , so you would be really helpful if you can teach me how to evaluate the integral

Answer 12

\[\sqrt{(t\sqrt{t^2+4)^2}+(-t \sqrt{4-t ^{2}})^2}\]
\[\sqrt{t^2(t^2+4)+t^2(4-t^2)}\]
\[\sqrt{t^2(t^2+4+4-t^2)}\]
\[\sqrt{8t^2}=2t\sqrt2\]

Answer 13

since its about the x axis, we need to define y from y' = t sqrt(4-t^2) right?

Answer 14

dropped a - on the y' ...

Answer 15

I've proven that y= \[\frac{ 1 }{ 3 } \left( 4-t ^{2} \right)^{\frac{ 3 }{ 2 }}\]

Answer 16

-t (4-t^2)^(1/2) i agree with that part; now its just :

\[2\pi\int\frac{ 1 }{ 3 } \left( 4-t ^{2} \right)^{\frac{ 3 }{ 2 }}~t\sqrt8~dt\]
\[\frac{2\sqrt8}{3}\pi\int t\left( 4-t ^{2} \right)^{\frac{ 3 }{ 2 }}~dt\]

which is pretty much just like integrating the y'

Answer 17

Got it! thx!

Answer 18

youre welcome

Answer 19

I'll be posting another question in just a few minutes, please take a look

Answer 20

if ive got the time ... need to make sure my own school work is attended to :)