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

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EyadBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
\[A = \int 2 \pi y \sqrt{(1 + [f '(y)]^2)} dy\] \[= (2 \pi) * \int_{0}^{6} y \sqrt{(1 + [2y/9]^2)}dy\] \[= \frac{2\pi}{9}* \int_{0}^{6} y \sqrt{(81 + 4y^2)} dy\] \[= \frac{2\pi}{9} * \left[\frac18 * (\frac23)(81 + 4y^2)^{\frac32} \right] _{0}^{6}= \frac{98\pi}{3}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
how did we go from line 2 to 3?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
We're rotating around the xaxis correct?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
In that case, you have \[y^2=9x18\implies y=\pm3\sqrt{x2}\]However, since we're looking for a full rotation, we throw out the negative. We want to integrate \[\large 3\pi\int_2^6\left(\sqrt{x2}\right)^2dx\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
oh shoot I'll do it again
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I think eyad might have been rotating around the yaxis and finding surface area.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Wait, we want surface area and not volume.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
In that case, we should be integrating \[\large 6\pi\int_2^6\left(\sqrt{x2}\right)dx\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I think you're right
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
\[2\pi r\]But we can factor a 3 out of the r.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
oh ok...lets see if I can do the integral
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
with my latex skills it's gonna be about 15 min just so you know
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I've got plenty of time.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
\[9x=y^2+18\] \[y=\sqrt{9x18}\] \[\frac{dy}{dx}=\frac 92 \left(9x18\right)^{\frac 12}\] \[S= \int_{2}^{6} 2 \pi x \sqrt{(1 + [f '(x)]^2)} dx\] \[S= \int_{2}^{6} 2 \pi x \sqrt{\left(1 + \left[\frac 92 \left(9x18\right)^{\frac 12}\right]^2\right)} dx\] \[S= 2 \pi\int_{2}^{6} x \sqrt{\left(1 + \left[\frac 92 \left(9x18\right)^{\frac 12}\right]^2\right)} dx\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
still working on it...
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
You're doing this using arc length?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I would not recommend that method.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
No I thought that was for surface area
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
what method would you recommend?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I'm soo silly ...sorry
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
We would only solve for x if we're revolving around the yaxis.
 one year ago
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