More Volume Problems — Question 5

Question 5

The base of a solid is the region bounded by $$y=x \quad\text{and}\quad y=x^2$$ between their intersection points. Cross-sections perpendicular to the $x$-axis are rectangles whose height is twice the base in the region. Find the volume of the solid.

Question 5 – Solution

1. Find the interval and upper curve.

$$x=x^2\quad\Longrightarrow\quad x(x-1)=0\quad\Longrightarrow\quad x=0,1.$$On $[0,1]$, $x\ge x^2$, so the line lies above the parabola.

2. Identify the rectangle dimensions.

The base is the vertical distance between curves; its height is twice that distance:$$b(x)=x-x^2,\qquad h(x)=2b(x)=2(x-x^2).$$

3. Find the area and expand.

$$A(x)=b(x)h(x)=2(x-x^2)^2=2x^2-4x^3+2x^4.$$

4. Set up and integrate the volume.

$$V=\int_0^1 A(x)\,dx=\left[\frac23x^3-x^4+\frac25x^5\right]_0^1.$$Each term follows from $\int x^n\,dx=x^{n+1}/(n+1)$.

5. Evaluate and combine the fractions.

$$V=\left(\frac23-1+\frac25\right)-0=\frac{10-15+6}{15}=\boxed{\frac1{15}}.$$

All volumes are in cubic units.