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# bounds on surface of revolution integral?

The question says:Set up an integral for the area of the surface obtained by rotating

the curve about y axis

Function: y=e^(-x^2) where x is equal to or between -1 and 1

What I did: integral(2pixds)

integral from -1 to 1 (2 * pi * x * square root ( 1 +4x^2 e^(-2x^2)

I got it wrong. The answer in the book says the bounds are 0 to 1. Everything else I did right.

Please tell me what i did wrong. I have a quiz tomorrow

so do you always do this when the bounds would be from a negative to positive? will the problems be "rigged" so that there is always symmetry when the bounds would be from a negative to positive?

sorry guys i'm still confused. is the back of the book wrong? there's no other two's in the answer key- it's exactly the same thing i got except for the bounds.

ok i get kb and mathmom's explanation i think

### 2 Answers

- MechEng2030Lv 77 years ago
You need to multiply your current integral by 2 and change the limits from 0 to 1 due to symmetry. Note that since the integral represents the area underneath a curve, the two areas would cancel out over here due to symmetry and lead to an answer of 0 had you kept the limits from -1 to 1.

- kbLv 77 years ago
Since we are rotating the region about the y-axis, the portion of the region on [-1,0] will be covered from rotating the region on [0, 1] about the y-axis. So, we only need to use 0 to 1 as bounds.

I hope this helps!