anonymous
  • anonymous
Find the area obtained from obtained by rotating the curve about the axis \[9x=y^2+18\] \[2 \le x \le 6\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[A = \int 2 \pi y \sqrt{(1 + [f '(y)]^2)} dy\] \[= (2 \pi) * \int_{0}^{6} y \sqrt{(1 + [2y/9]^2)}dy\] \[= \frac{2\pi}{9}* \int_{0}^{6} y \sqrt{(81 + 4y^2)} dy\] \[= \frac{2\pi}{9} * \left[\frac18 * (\frac23)(81 + 4y^2)^{\frac32} \right] _{0}^{6}= \frac{98\pi}{3}\]
anonymous
  • anonymous
how did we go from line 2 to 3?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.