help lagrangeson678 with "find the surface area swept over by the portion of the parabola 9y^2=4x lying between the lines x=0 and x=1 when revolves through 2 right angles about the x-axis

Question

amistre64 · Answer

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amistre64 · Answer

and when it revolves at 2 rt angles?  aint that 180 degress?

amistre64 · Answer

id say add up all the circumferences of the circles from 0 to 1 with a modified function as your radius

amistre64 · Answer

does that make sense?

anonymous · Answer

not really, where did you get circles from

amistre64 · Answer

when you stare into the inside of a bowl, its just a bunch of circles held together

anonymous · Answer

can you please help me work the steps for this problem, i really need some help

amistre64 · Answer

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amistre64 · Answer

$$\int_{0}^{1}\ 2\pi\ [f(x)]dx$$
$$\left. \pi\ [f(x)]^2\ \right]_{0}^{1}$$
$$\pi\ [f(1)]^2\ - \pi\ [f(0)]^2$$

amistre64 · Answer

we know 9y^2 = 4x ; so we solve for y.

y = sqrt[(4/9)x]
   = (2/3) sqrt(x)

$$\pi\ [f(1)]^2\ - \pi\ [f(0)]^2$$
$$\pi\ [(2/3) \sqrt(1)]^2\ - \pi\ [(2/3) \sqrt(0)]^2$$
$$=\frac{4\ pi}{9}$$
but lets check the wolfram :)

amistre64 · Answer

8pi/9 is the wolf   means i mis typed it in here someplace

amistre64 · Answer

$$\frac{4pi}{3}\int_{0}^{1}\ \sqrt{x}\ dx$$
$$\frac{4pi}{3} \frac{2}{3}\sqrt{x^3}\ $$
$$\frac{8pi}{9} $$ thats as i see it

anonymous · Answer

how do you get 4pi/3?

amistre64 · Answer

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