@TuringTest and @SmoothMath does that mean that I can write $$\int_{2}^{6} 2\pi x \sqrt{1+{(\frac{dy}{dx})}^2}dx$$
as
$$\int_{2}^{6} 2\pi x \sqrt{1+{(\frac{dx}{dy})}^2}dx$$
for $$9x=y^2+18$$ and 2

Question

@TuringTest and @SmoothMath does that mean that I can write $$\int_{2}^{6} 2\pi x \sqrt{1+{(\frac{dy}{dx})}^2}dx$$
as
$$\int_{2}^{6} 2\pi x \sqrt{1+{(\frac{dx}{dy})}^2}dx$$
for $$9x=y^2+18$$ and 2<=x<=6

anonymous · Answer

$$x=y^2 /9 +2$$

anonymous · Answer

x=(y^2 /9) +2

turingtest · Answer

you have to be careful with the bounds
if you integrate with respect to y instead of x you need to find the new bounds on y

anonymous · Answer

I'll be right back

turingtest · Answer

me too

anonymous · Answer

oh ok, so If it's x=(y^2 /9) +2 I would have to change the bounds to
$$y=\sqrt{9(2)-18}=0$$
$$y=\sqrt{9(6)-18}=6$$

turingtest · Answer

right

anonymous · Answer

ok

turingtest · Answer

other than that though the answer to you question is "yes"
(and note that the second integral you wrote should be wrt y, not x)

turingtest · Answer

your*

anonymous · Answer

so if it's wrt y is it dy at the end?

anonymous · Answer

I just googled wrt y...and yes it's dy

turingtest · Answer

yeah sorry I don't reply immediately
multitasking...

anonymous · Answer

Same here....I'm actually on the clock till 11:30pm :P 
" The first part has the form 2pi*something, where that something is the radius."  according to SmoothMath so that should be the same whether I integrate wrt y or wrt x...correct?

turingtest · Answer

yes, you can always chose whether to integrate wrt y or x in these problems involving arc length and surface area
again, just be sure to change the bounds

SmoothMath's way is a good way to think about it; adding up an infinity of circumferences    with radii dependent on where you are on the graph.

anonymous · Answer

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