Question

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

Answer 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.

Answer 2

\[A = \int 2 \pi y \sqrt{(1 + [f '(y)]^2)} dy\]
\[= (2 \pi) * \int_{0}^{6} y \sqrt{(1 + [2y/9]^2)}dy\]
\[= \frac{2\pi}{9}* \int_{0}^{6} y \sqrt{(81 + 4y^2)} dy\]
\[= \frac{2\pi}{9} * \left[\frac18 * (\frac23)(81 + 4y^2)^{\frac32} \right] _{0}^{6}= \frac{98\pi}{3}\]

Answer 3

how did we go from line 2 to 3?

Answer 4

I get the 4y^2

Answer 5

but not the 81

Answer 6

We're rotating around the x-axis correct?

Answer 7

yes
I believe so

Answer 8

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 \[\large 3\pi\int_2^6\left(\sqrt{x-2}\right)^2dx\]

Answer 9

did I do it wrong?

Answer 10

oh shoot I'll do it again

Answer 11

I think eyad might have been rotating around the y-axis and finding surface area.

Answer 12

Wait, we want surface area and not volume.

Answer 13

In that case, we should be integrating \[\large 6\pi\int_2^6\left(\sqrt{x-2}\right)dx\]

Answer 14

I think you're right

Answer 15

why 6 pi

Answer 16

\[2\pi r\]But we can factor a 3 out of the r.

Answer 17

oh ok...lets see if I can do the integral

Answer 18

with my latex skills it's gonna be about 15 min just so you know

Answer 19

I've got plenty of time.

Answer 20

\[9x=y^2+18\]
\[y=\sqrt{9x-18}\]
\[\frac{dy}{dx}=\frac 92 \left(9x-18\right)^{-\frac 12}\]
\[S= \int_{2}^{6} 2 \pi x \sqrt{(1 + [f '(x)]^2)} dx\]
\[S= \int_{2}^{6} 2 \pi x \sqrt{\left(1 + \left[\frac 92 \left(9x-18\right)^{-\frac 12}\right]^2\right)} dx\]
\[S= 2 \pi\int_{2}^{6} x \sqrt{\left(1 + \left[\frac 92 \left(9x-18\right)^{-\frac 12}\right]^2\right)} dx\]

Answer 21

still working on it...

Answer 22

You're doing this using arc length?

Answer 23

I would not recommend that method.

Answer 24

No I thought that was for surface area

Answer 25

what method would you recommend?

Answer 26

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.

Answer 27

I'm soo silly ...sorry

Answer 28

We would only solve for x if we're revolving around the y-axis.