More Volume Problems — Question 10

Question 10

The base of a solid is the region bounded by $$y=\sqrt{x} \quad\text{and}\quad y=x^2,$$ between their points of intersection. Cross-sections parallel to the $y$-axis are rectangles whose width is equal to the distance between the curves and whose height is constant and equal to $2$. Find the volume of the solid.

Question 10 – Solution

1. Interpret the slice direction and find the bounds.

The base segments are parallel to the $y$-axis, so they are vertical; integrate with respect to $x$. Solving $\sqrt{x}=x^2$ gives $x=x^4$, or $x(x^3-1)=0$. The real intersections occur at $x=0,1$.

2. Identify width and height.

For $0\le x\le1$, $\sqrt{x}\ge x^2$. Thus each rectangle has$$w(x)=\sqrt{x}-x^2,\qquad h=2.$$The height of $2$ extends out of the base plane.

3. Find the area and set up the integral.

$$A(x)=2(\sqrt{x}-x^2),\qquad V=\int_0^1 A(x)\,dx=2\int_0^1(x^{1/2}-x^2)dx.$$

4. Integrate using the power rule.

$$\int x^{1/2}\,dx=\frac23x^{3/2},\qquad\int x^2\,dx=\frac{x^3}{3}.$$$$V=2\left[\frac23x^{3/2}-\frac{x^3}{3}\right]_0^1.$$

5. Apply the bounds and simplify.

$$V=2\left[\left(\frac23-\frac13\right)-0\right]=\boxed{\frac23}.$$

All volumes are in cubic units.