## MathSofiya 3 years ago Find the area obtained from obtained by rotating the curve about the axis $9x=y^2+18$ $2 \le x \le 6$

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. MathSofiya

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

3. MathSofiya

how did we go from line 2 to 3?

4. MathSofiya

I get the 4y^2

5. MathSofiya

but not the 81

6. KingGeorge

We're rotating around the x-axis correct?

7. MathSofiya

yes I believe so

8. 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 $\large 3\pi\int_2^6\left(\sqrt{x-2}\right)^2dx$

9. MathSofiya

did I do it wrong?

10. MathSofiya

oh shoot I'll do it again

11. KingGeorge

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

12. KingGeorge

Wait, we want surface area and not volume.

13. KingGeorge

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

14. MathSofiya

I think you're right

15. MathSofiya

why 6 pi

16. KingGeorge

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

17. MathSofiya

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

18. MathSofiya

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

19. KingGeorge

I've got plenty of time.

20. MathSofiya

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

21. MathSofiya

still working on it...

22. KingGeorge

You're doing this using arc length?

23. KingGeorge

I would not recommend that method.

24. MathSofiya

No I thought that was for surface area

25. MathSofiya

what method would you recommend?

26. 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.

27. MathSofiya

I'm soo silly ...sorry

28. KingGeorge

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