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# Calculus/area of the surface when revolved

Question
find the area of the surface generated when the given curve is revolved about the x-axis
y=12x-2 between (5/12,3) and (13/12,11)

The area at position x is πy² where y is given as (12x-2)².
This means we need to integrate π(12x-2)² from x=5/12 to x=13/12.

The derivative of 12x-2 is 12, so the integral is π(12x-2)³/(3*12) = π(12x-2)³/36.

Evaluating that from 13/12 down to 5/12 gives π(13-2)³/36 - π(5-2)³/36.
It is known that 13-2=11, and 11*11*11 = 1,331.  It is also known that 5-2 = 3 and 3*3*3 = 27.
If we take 1,331 - 27 we get 1,304.

From here, it means the answer is (1,304/36)π = (326/9)π.
Well, 36*9 = 324, so there are 2/9 left over, so I get (36 2/9)π.

Now I could have squared 12x-2 and gotten 144x² - 48x + 4, but that leads to a lot more calculations.

If we did this, integrating would have given 48x³ - 24x² + 4x.

Since 13³/12³ = 2,197/144 and 5³/12³ = 125/1,728, the difference is 2,072/1,728.
Note that if we take this times 48, we get 2,072/36.
Dividing the numerator and the denominator by 4 gives 518/9.

Since 13²/12² = 169/144 and 5²/12² = 25/144, the difference is 144/144 = 1.
Note that if we take this times -24 we get -24.

For the last term, 13/12 - 5/12 = 8/12 = 2/3.  Note that if we take this times 4 gives 8/3.

Adding all the terms together gives us 326/9,
which when we remember to multiply by π gives us the same thing.

From seeing this, it can be seen that multiplying out before integration can lead to a harder problem.
This means to do it only if necessary.

Calculus

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