The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross...

Question:

The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x{eq}^2 {/eq}. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

Volumes of Solids

The volume of a solid whose cross section is {eq}\displaystyle A(x), {/eq} for {eq}\displaystyle x\in[a,b] {/eq} is given as {eq}\displaystyle \int_a^b A(x)\ dx. {/eq}

If the cross sections are equilateral triangles with the side {eq}\displaystyle l=l(x), {/eq} then the area of a equilateral triangle is {eq}\displaystyle \frac{l^2\sqrt{3}}{4}, {/eq}

Answer and Explanation:

The solid with the base, the region bounded by the curves {eq}\displaystyle y=x^2, y=x {/eq} and the cross sections, equilateral triangles,

have the side of the triangles of length l equal to the vertical distance of the region, which is {eq}\displaystyle l= x-x^2, {/eq} for {eq}\displaystyle 0\leq x\leq 1. {/eq}

We obtained the range of x by determining the points of intersection of the line with parabola: {eq}\displaystyle y=x=x^2\iff x(x-1)=0\implies x=0, x=1. {/eq}

Therefore, the area is {eq}\displaystyle A(x)= \frac{(x-x^2)^2\sqrt{3}}{4}. {/eq}

and the volume is

{eq}\displaystyle \begin{align}\int_0^{1} \mathcal{A}(x)\ dx &=\frac{\sqrt{2}}{4}\int_0^{1} (x-x^2)^2\, dx\\ &=\frac{\sqrt{3}}{4}\int_0^{1} \left(x^2-2x^3+x^4\right)\, dx\\ &=\frac{\sqrt{3}}{4} \left(\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right)\bigg\vert_0^1\\ &=\frac{\sqrt{3}}{4} \left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)\\ &=\boxed{\frac{\sqrt{3}}{120} }. \end{align} {/eq}


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