Question

Can someone help me Find the lateral (side) surface area of the cone generated by revolving the line segment y = x/2, 0 <= x <= 3, about the x-axis.

Answer 1

Any help is appreciated

Answer 2

What have you tried on this problem so far? Any ideas?

Answer 3

We were covering rotation of x-axis problems of integrals but I've had the flu and missed class, thus I'm left to try to figure this out sadly solo so no sorry :( I'd appreciate the guidance/steps though so I can understand!

Answer 4

This might be a good resource to view: http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
We will need two particular formulas from it.
\( \displaystyle S = \int_{a}^{b} 2 \pi y \ ds \) Revolving about x-axis.
\( ds = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} dx \)
Intuitively, think of surface area as if we are trying to find circumference of circles of radius y and giving it a tiny amount of area multiplying by a width ds along the curve. The integral sums up all these tiny areas to give us the total surface area.

Answer 5

so we use the x/2 within the formula you provided such as: integral upper as 3, lower as 0, then 2(pi)(x/2) y?

Answer 6

integral from 0 to 3 of 2 pi (x/2) is good (no y), then we have that extra ds.
The ds is essentially the tiniest width that is in the direction of y=x/2. By Pythagorean Thm: ds^2 = dy^2 + dx^2 where dy goes along the y-values and dx goes along the x-values.
We convert it into terms of dx to make the integral able to be evaluated.

Answer 7

Or in symbols: \( \displaystyle \int_{0}^{3} 2 \pi \left(\dfrac{x}{2} \right) \ \color{red}{ds} \)

Answer 8

i follow the formula but where do we gather the dy^2 and dx^2 from?

Answer 9

That is more of a definition than anything. ds is a tiny length along the curve, while dx and dy are along their respective axes.