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A solid is bounded by two bases in the horizontal planes $z = h/2$ and $z = -h/2,$ and by such a surface that the area of every section in a horizontal plane is given by a formula of the sort $$\text{Area} = a_0z^3+a_1z^2+a_2z+a_3$$ (where as special cases some of the coefficients may be $0$). Show that the volume is given by the formula $$\dfrac{1}{6}h[B_1+B_2+4M],$$ where $B_1$ and $B_2$ are the areas of the bases, and $M$ is the area of the middle horizontal section.

Attempt

We take the $V = \displaystyle \int_{-h/2}^{h/2} (a_0z^3+a_1z^2+a_2z+a_3) = \dfrac{a_1h^3}{12}+a_3h$. Then we have to show this equals $\dfrac{1}{6}h[B_1+B_2+4M].$ Since $B_1 = a_0(-h/2)^3+a_1(-h/2)^2+a_2(-h/2)+a_3$, $B_2 = a_0(h/2)^3+a_1(h/2)^2+a_2(h/2)+a_3$, and $M = a_0(0)^3+a_1(0)^2+a_2(0)+a_3(0)=0$, we plug them into the formula and get, $$\dfrac{1}{6}h[B_1+B_2+4M] = \dfrac{1}{6}h[(a_0(-h/2)^3+a_1(-h/2)^2+a_2(-h/2)+a_3)+(a_0(h/2)^3+a_1(h/2)^2+a_2(h/2)+a_3)+4 \cdot 0] \neq \dfrac{a_1h^3}{12}+a_3h \hspace{6 mm}\text{if my calculations are right}.$$

Therefore I am confused where I went wrong.

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  • $\begingroup$ Note that this exercise shows that Simpson's Rule is exact for cubic functions. $\endgroup$
    – David K
    Jan 3, 2016 at 4:24

1 Answer 1

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You should have $M=a_3$, since the last term in the area isn't multiplied by $z$. Then your calculation will give the correct answer.

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