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yea giving it bounds would be the only way to do this or else the graph crosses former self when you rotate.
i contend that since you're rotating about the x axis that ybar = 0. due to symmetry.
also, zbar = 0 also because of the symmetry that you get from rotating about the x axis.
so the only thing that matters is xbar which is simply the x value of the average value of e^(-x) between 0 and 1.
so the average value of e^(-x) would be 1/(1-0) * int[0,1][e^(-x)dx] and you get -1/e - (-1) = 1-1/e or (e-1)/e.
to get the x value of that you say that e^(-x) = (e-1)/e, take the ln() of both sides and multiply by -1 to get x.
that leaves -ln[(e-1)/e] or ln[e/(e-1)]
which becomes 1- ln[e-1]. what ever that value is should be your xbar.
xbar = 1-ln[e-1].
ybar = 0
zbar = 0
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