More Volume Problems — Question 4

Question 4

The base of a solid is the region enclosed by $$y=1-x^2 \quad\text{and}\quad y=0$$ between their intersection points. Cross-sections perpendicular to the $x$-axis are equilateral triangles. Find the volume of the solid.

Question 4 – Solution

1. Find the bounds and side length.

The intersections satisfy $1-x^2=0$, so $x=-1,1$. A vertical base segment gives$$s(x)=1-x^2\qquad(-1\le x\le1).$$

2. Find the area of an equilateral triangle.

The altitude is $\sqrt{s^2-(s/2)^2}=\sqrt3\,s/2$. Hence$$A=\frac12s\left(\frac{\sqrt3}{2}s\right)=\frac{\sqrt3}{4}s^2,\qquad A(x)=\frac{\sqrt3}{4}(1-x^2)^2.$$

3. Set up the volume and expand.

$$V=\frac{\sqrt3}{4}\int_{-1}^1(1-x^2)^2dx=\frac{\sqrt3}{4}\int_{-1}^1(1-2x^2+x^4)dx.$$

4. Use symmetry and integrate.

The integrand is even, so$$V=\frac{\sqrt3}{2}\int_0^1(1-2x^2+x^4)dx=\frac{\sqrt3}{2}\left[x-\frac{2x^3}{3}+\frac{x^5}{5}\right]_0^1.$$

5. Evaluate and simplify.

$$V=\frac{\sqrt3}{2}\left(1-\frac23+\frac15\right)=\frac{\sqrt3}{2}\left(\frac{15-10+3}{15}\right)=\boxed{\frac{4\sqrt3}{15}}.$$

All volumes are in cubic units.