Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
MathSofiya
Group Title
Find the area obtained from obtained by rotating the curve about the axis
\[9x=y^2+18\]
\[2 \le x \le 6\]
 2 years ago
 2 years ago
MathSofiya Group Title
Find the area obtained from obtained by rotating the curve about the axis \[9x=y^2+18\] \[2 \le x \le 6\]
 2 years ago
 2 years ago

This Question is Closed

Eyad Group TitleBest 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.
 2 years ago

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
how did we go from line 2 to 3?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
I get the 4y^2
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
but not the 81
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
We're rotating around the xaxis correct?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
yes I believe so
 2 years ago

KingGeorge Group TitleBest 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 \[\large 3\pi\int_2^6\left(\sqrt{x2}\right)^2dx\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
did I do it wrong?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
oh shoot I'll do it again
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I think eyad might have been rotating around the yaxis and finding surface area.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Wait, we want surface area and not volume.
 2 years ago

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
I think you're right
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
why 6 pi
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
\[2\pi r\]But we can factor a 3 out of the r.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
oh ok...lets see if I can do the integral
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
with my latex skills it's gonna be about 15 min just so you know
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I've got plenty of time.
 2 years ago

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
still working on it...
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
You're doing this using arc length?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I would not recommend that method.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
No I thought that was for surface area
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
what method would you recommend?
 2 years ago

KingGeorge Group TitleBest 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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.0
I'm soo silly ...sorry
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
We would only solve for x if we're revolving around the yaxis.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.