Finding volume (again)... is this correct?

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fbellman

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Jan 8, 2006
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Hi all,

So this is the end of my posting spree. I'm on the last part of the last problem, thank goodness. This relates to my last post (where I asked how to take the integral of y=e^(-x^2)... I ended up just using a calculator, like someone recommended! Thanks!).

So, here's the full problem:
Let R be the shaded region in the first quadrant enclosed by the graphs of y=e^(-x^2) and y=(1-cos x) and the y-axis.

A) Find the area of the region R. I did this by taking the integral from 0 to 1 [e^(-x^2)-1-cos x]dx. I got .5883

B) Find the volume of the solid generated when the region R is revolved about the x-axis. I did this one by taking pi times the integral from 0 to 1 of [(e^(-x^2))^2 - (1-cos x)^2]dx. My answer was .554*pi.

C) The region r is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid.

^^ This is where I'm stuck. I tried to follow my (awful) textbook, and did it like so:

Integral from 0 to 1 of [2(e^(-x^2)- 1 + cos x)]^2*dx.

I don't think this is correct because when I graphed it... well, it doesn't really fall in the interval of 0 to 1...

Questions: Are parts A and B correct? How should I set up the equation for part C?

Thanks a lot!!
 
First off, make sure you have the function graphed.

a. You have the integrand correct. 0 is a correct limit of integration, but why is the second limit 1?

You need the point of intersection of the two curves, so e^(-x^2)=1-cos(x)
Using a calculator you should get x=0.941944=0.942

Don't round this up to 1! You will lose points on the AP exam for doing so (If you are taking an AP course. Otherwise, your answer is simply less accurate)

So evaluating from 0 to 0.942 should get you approximately 0.591.

b. Again, everything is correct but the limits. Change 1 to 0.942.

c. A side of the square is e^(-x^2)-(1-cos(x)), and the area of a square with side x is x^2, so you just evaluate the integral from 0 to 0.942 of [e^(-x^2)-(1-cos(x))]^2 dx
 
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