Why is it that the surface of revolution integral contains the arc lengh bit

http://upload.wikimedia.org/math/8/9...470bd9a390.png

but the volume of revolution doesnt?

http://upload.wikimedia.org/math/4/8...61d4e2e830.png

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- Oct 20th 2010, 09:01 AMfloatersurface/volume integration
Why is it that the surface of revolution integral contains the arc lengh bit

http://upload.wikimedia.org/math/8/9...470bd9a390.png

but the volume of revolution doesnt?

http://upload.wikimedia.org/math/4/8...61d4e2e830.png - Oct 22nd 2010, 02:50 AMmr fantastic
- Oct 22nd 2010, 07:02 AMHallsofIvy
When you calculate volume, you are approximating the area that is rotated around the axis by a thin rectangle and can use the y value at any point to approximate the height of the rectangle. But when you calculate surface area, it is the

**slant**height that has to be rotated around the axis, not the horizontal length of each small section.

Think about this:

If you take the line from (0, 0) to (1, 1), y= x, you can approximate the area under it taking thin rectangles with width $\displaystyle \Delta x$ and height y= x for x any value in that interval. The total area is [tex]\sum x_i \Delta x[tex] that in the limit becomes $\displaystyle \int x dx$ which gives the correct area.

But if you attempt to approximate the**length**of the line by using little horizontal sections, the length will always total to 1, not the correct length, no matter how many sections you take. To get the correct length, from $\displaystyle x_i$ to $\displaystyle x_i+ \Delta x$, you have to use the actual length of the line, $\displaystyle \sqrt{(x_i+ \Delta x- x_i)^2+ (y_{i+1}- y_i)^2}= \sqrt{x_i^2+ x_i}= x_i\sqrt{2}$. - Oct 22nd 2010, 11:54 PMfloater
Thankyou. I guess I thought that because dx is infinitesimally then the error wouldn't exist. But I now realise how that wouldnt work for anything other than a straight line.

Thnakyou