What is the volume of the solid given the base of a solid is the region in the first quadrant bounded by the graph of #y=-x^2+5x-4^ and the x-axis and the cross-sections of the solid perpendicular to the x-axis are equilateral triangles?

1 Answer
Mar 22, 2015

The volume is \frac{81\sqrt{3}}{40}\approx 3.5074.

The curve y=-x^2+5x-4=-(x-1)(x-4) has x-intercepts at x=1 and x=4 (and is above the x-axis for 1<x<4).

For 1< x< 4, let b(x)=-x^2+5x-4. This will be the base of the equilateral triangle cross-section at x.

The solid itself looks something like this:

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Now draw the cross-section equilateral triangle, label the sides b(x), and draw a vertical line for the height h(x) from the top vertex down perpendicular to the base.
By the Pythagorean Theorem, (b(x))^2=(\frac{b(x)}{2})^2+(h(x))^2.

enter image source here

so that h(x)=\sqrt{(b(x))^2-(\frac{b(x)}{2})^2}=\sqrt{\frac{3}{4}(b(x))^2}=\frac{\sqrt{3}}{2}b(x).

The cross-sectional area is A(x)=\frac{1}{2}b(x)h(x)=\frac{\sqrt{3}}{4}(b(x))^2=\frac{\sqrt{3}}{4}(x^{4}-10x^{3}+33x^{2}-40x+16), which means the volume of the solid is V=\int_{1}^{4}A(x)dx=\frac{\sqrt{3}}{4}\int_{1}^{4}(x^{4}-10x^{3}+33x^{2}-40x+16)dx=\frac{\sqrt{3}}{4}(\frac{x^{5}}{5}-\frac{5x^{4}}{2}+11x^{3}-20x^{2}+16x)|_{x=1}^{x=4}.

Continuing to simplify gives V=\frac{\sqrt{3}}{4}((\frac{1024}{5}-\frac{1280}{2}+704-320+64)-(\frac{1}{5}-\frac{5}{2}+11-20+16))=\frac{\sqrt{3}}{4}(\frac{64}{5}-\frac{47}{10})=\frac{81\sqrt{3}}{40}.