area of surface formed by rotating y=x^2, 0<=x<=1, abt x-axi

[MarkSA]

Hello,

I have the question:
1) Find the area of the surface obtained by rotating the curve about the x-axis:
y=x^2, 0<=x<=1

I'm having some trouble integrating this one. I've tried both ways - putting the equation in terms of x and y.
If I put it in terms of y I get:
x = sqrt(y)
S = 2pi*integral from 0 to 1 of: y * sqrt[1 + 1/(4y)] dy

If I leave it in terms of x I get:
y = x^2
S = 2pi*integral from 0 to 1 of: x^2 * sqrt(1 + 4x^2) dx

Is there a fairly simple way of integrating these i'm missing? Thanks

 

[soroban]

Re: Find the area of the surface

Hello, Mark!

1) Find the area of the surface obtained by rotating the curve about the x-axis:
. . \(\displaystyle y\:=\:x^2,\quad 0\leq x \leq 1\)

If I leave it in terms of \(\displaystyle x\) I get:
. . \(\displaystyle S \;=\; 2\pi \int^1_0 x^2\sqrt{1 + 4x^2}\,dx\) . . . . Good!

I would use Trig Substitution . . .

\(\displaystyle \text{Let }\,2x\:=\:\tan\theta\quad\Rightarrow\quad x \:=\:\frac{1}{2}\tan\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{2}\sec^2\!\theta\,d\theta\)

. . \(\displaystyle \text{and: }\;\sqrt{1+4x^2} \;=\;\sqrt{1 + \tan^2\!\theta} \;=\;\sqrt{\sec^2\!\theta} \;=\;\sec\theta\)


\(\displaystyle \text{Substitute: }\;S \;=\;2\pi\int \frac{1}{4}\tan^2\!\theta\cdot\sec\theta\left(\frac{1}{2}\sec^2\!\theta\,d\theta\right) \;=\;\frac{1}{4}\pi}\int\sec^3\!\theta\tan^2\!\theta\,d\theta\)

. . . . . . Good luck!

 

[MarkSA]

Re: Find the area of the surface

thank you