Find the area obtained from obtained by rotating the curve about the axis \[9x=y^2+18\] \[2 \le x \le 6\]

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Find the area obtained from obtained by rotating the curve about the axis \[9x=y^2+18\] \[2 \le x \le 6\]

Mathematics
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Write the curve as x = y^2/9 + 2. If x = 2, then y = 0 If x = 6, then y = 6 (the rotation of this region takes care of the negative counterpart.) A = integral 2 pi y sqrt(1 + [f '(y)]^2) dy = (2 pi) * integral(y = 0 to 6) y sqrt(1 + [2y/9]^2) dy = (2 pi/9) * integral(y = 0 to 6) y sqrt(81 + 4y^2) dy = (2 pi/9) * [(1/8) * (2/3)(81 + 4y^2)^(3/2) for y = 0 to 6] = 98 pi / 3.
\[A = \int 2 \pi y \sqrt{(1 + [f '(y)]^2)} dy\] \[= (2 \pi) * \int_{0}^{6} y \sqrt{(1 + [2y/9]^2)}dy\] \[= \frac{2\pi}{9}* \int_{0}^{6} y \sqrt{(81 + 4y^2)} dy\] \[= \frac{2\pi}{9} * \left[\frac18 * (\frac23)(81 + 4y^2)^{\frac32} \right] _{0}^{6}= \frac{98\pi}{3}\]
how did we go from line 2 to 3?

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I get the 4y^2
but not the 81
We're rotating around the x-axis correct?
yes I believe so
In that case, you have \[y^2=9x-18\implies y=\pm3\sqrt{x-2}\]However, since we're looking for a full rotation, we throw out the negative. We want to integrate \[\large 3\pi\int_2^6\left(\sqrt{x-2}\right)^2dx\]
did I do it wrong?
oh shoot I'll do it again
I think eyad might have been rotating around the y-axis and finding surface area.
Wait, we want surface area and not volume.
In that case, we should be integrating \[\large 6\pi\int_2^6\left(\sqrt{x-2}\right)dx\]
I think you're right
why 6 pi
\[2\pi r\]But we can factor a 3 out of the r.
oh ok...lets see if I can do the integral
with my latex skills it's gonna be about 15 min just so you know
I've got plenty of time.
\[9x=y^2+18\] \[y=\sqrt{9x-18}\] \[\frac{dy}{dx}=\frac 92 \left(9x-18\right)^{-\frac 12}\] \[S= \int_{2}^{6} 2 \pi x \sqrt{(1 + [f '(x)]^2)} dx\] \[S= \int_{2}^{6} 2 \pi x \sqrt{\left(1 + \left[\frac 92 \left(9x-18\right)^{-\frac 12}\right]^2\right)} dx\] \[S= 2 \pi\int_{2}^{6} x \sqrt{\left(1 + \left[\frac 92 \left(9x-18\right)^{-\frac 12}\right]^2\right)} dx\]
still working on it...
You're doing this using arc length?
I would not recommend that method.
No I thought that was for surface area
what method would you recommend?
Just use the formula for circumference of a circle. \(C=2\pi r\). Where \(r\) is your function solved for \(y\). Integrate from 2 to 6, and bam! There's the answer.
I'm soo silly ...sorry
We would only solve for x if we're revolving around the y-axis.

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