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this problem is on page 267 of Advanced calculus of several variables by Edwards, I just can't seem to get a handle on it:

Let $A$ be a contented set in the right half of the $xz$-plane $x>0$. Define $\hat{x}$, the $x$-coordinate of the centroid of A, by $\hat{x}=[1/v(A)]\int\int_Axdxdz$. If $C$ is the set obtained by revolving about the z-axis, that is,$$C=\{(x,y,z)\in R^3:((x^2+y^2)^{1/2},z)\in A)\} $$

then Pappus' theorem asserts that $$v(C) = 2\pi\hat{x}v(A)$$

that is, that the volume of C is the volume of A multiplied by the distance traveled by the centroid of A. Note that C is the image under the cylindrical coordinates map of the set $$B = \{(r,\theta,z)\in R^3:(r,z) \in A, \theta \in [0,2\pi]\}$$

Apply the change of variables and iterated integrals theorems to to deduce Pappus' theorem

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ugh, i think i have made some headway, but i still cant get the answer, I am confused on how to get the Volume of A... is that by double integrating C ? –  Neo Apr 26 '13 at 11:35

1 Answer 1

up vote 1 down vote accepted

When you are dealing with change of variables, find the Jacobian determinant. We know that $x=r\cos\theta$, $y=r\sin\theta$, and $z=z$ in cylindrical coordinates. Thus, first find out that a change in variables would give you $rdrd\theta$ in the new integral.

Now, we know $v(C)=\iiint 1$ over the region $C$. With our change in variables, we get:

$$v(C)=\iint_A\int_0^{2\pi} rd\theta drdz$$ Thus, we now have $$v(C) = \int_c^d\int_a^b\int_0^{2\pi} rd\theta drdz$$ which turns into: $$2\pi\int_c^d\int_a^b rdrdz$$

Realize that after integrating $r$ to get $\frac{1}{2}r^2$ we get $\pi(b^2-a^2)$. This is the $v(A)$ we wanted because $b^2-a^2$ represents $[f(z)]^2$. So now our equation looks like this:

$$v(C)=\int_c^dv(A)\quad\mbox{---- Remember this because we use it later}$$

Now we solve the integral given to us for $\hat x$.

$$\hat x = \frac{1}{v(A)} \iint xdxdz = \frac{1}{v(A)} \int_c^d\int_a^b xdxdz = \frac{1}{2\pi*v(A)} \int_c^d v(A)$$ where we divide by $\pi$ in order to get the integral in terms of $v(A)$ because we know that:

$$b^2-a^2 = [f(z)]^2 = \frac{v(A)}{\pi}$$ because $\pi[f(z)]^2 = v(A)$

Also, know that the $\frac{1}{2}$ came from integrating $x$ to get $\frac{1}{2}x^2$

The familiar expression we got for $\hat x$ has $v(C)$ in it.

Meaning, this is the same as saying $\hat x = \frac{1}{v(A)}\frac{1}{2\pi}v(C)$

Multiplying both sides of the equation by $v(A)$ and $2\pi$ we get what we were looking for:

$$v(C)=2\pi\hat x v(A)$$

Hence, we are finished and the proof is concluded!

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this makes sense. thank you! –  Neo Apr 26 '13 at 11:55
    
For some basic information about writing math at this site see e.g. here, here, here and here. –  Lord_Farin Apr 26 '13 at 12:18

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