## 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

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 Group Title

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

3. MathSofiya Group Title

how did we go from line 2 to 3?

4. MathSofiya Group Title

I get the 4y^2

5. MathSofiya Group Title

but not the 81

6. KingGeorge Group Title

We're rotating around the x-axis correct?

7. MathSofiya Group Title

yes I believe so

8. KingGeorge Group Title

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 Group Title

did I do it wrong?

10. MathSofiya Group Title

oh shoot I'll do it again

11. KingGeorge Group Title

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

12. KingGeorge Group Title

Wait, we want surface area and not volume.

13. KingGeorge Group Title

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

14. MathSofiya Group Title

I think you're right

15. MathSofiya Group Title

why 6 pi

16. KingGeorge Group Title

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

17. MathSofiya Group Title

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

18. MathSofiya Group Title

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

19. KingGeorge Group Title

I've got plenty of time.

20. MathSofiya Group Title

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

21. MathSofiya Group Title

still working on it...

22. KingGeorge Group Title

You're doing this using arc length?

23. KingGeorge Group Title

I would not recommend that method.

24. MathSofiya Group Title

No I thought that was for surface area

25. MathSofiya Group Title

what method would you recommend?

26. KingGeorge Group Title

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 Group Title

I'm soo silly ...sorry

28. KingGeorge Group Title

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