Question

find the exact area of the surface obtained by rotating the given curve about the x-axis.

Answer 1

x = cos^3 t y= sin^3 t 0<theta<pi/2

Answer 2

hello

Answer 3

@Loser66 can you help me with this problem

Answer 4

i am confused as to what to do

Answer 5

@freckles

Answer 6

we want to use this
\[\int\limits_{}^{} 2 \pi y \sqrt{(x')^2+(y')^2} dt \]

Answer 7

alright

Answer 8

\[\int\limits_{0}^{\frac{\pi}{2}}2\pi \sin^3(t) \sqrt{((\cos^3(t))')^2+((\sin^3(t))')^2} dt\]

Answer 9

we need to differentiate the cos cos^3(t) and the sin^3(t) thingy w.r.t to t

Answer 10

for example
\[\frac{d}{dt} \sin^3(t)=3\sin^2(t) \cdot \cos(t)\]

then square both sides to find the square of the derivative since our formula calls for that:
\[(\frac{d}{dt} \sin^3(t))^2=(3 \sin^2(t) \cos(t))^2 =9 \sin^4(t) \cos^2(t)\]

so you try both of these things for the cos^3(t) part

Answer 11

this is what i have so far

Answer 12

\[\frac{d}{dt}\cos^3(t)=3\cos^2(t)(-\sin(t)) \\ (\frac{d}{dt} \cos^3(t))^2=9\cos^4(t)\sin^2(t)\]
so let's see we are going to put this in
and it looks like you have it ...
\[\int\limits_{0}^{\frac{\pi}{2}} 2\pi \sin^3(t) \sqrt{9\cos^4(t)\sin^2(t)+9\sin^4(t)\cos^2(t)} dt\]
well except that -1 part

Answer 13

(-1)^2=1

Answer 14

now we could factor inside that square root part

Answer 15

how do we do that