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Calculus IChapter 6: Application of IntegralsSection 6-5

More Volume Problems — Question 10

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Question 10

The base of a solid is the region bounded by y=xandy=x2,y=\sqrt{x} \quad\text{and}\quad y=x^2, between their points of intersection. Cross-sections parallel to the yy-axis are rectangles whose width is equal to the distance between the curves and whose height is constant and equal to 22. Find the volume of the solid.

See the diagram in the original worksheet below.

Shaded base region in the xyxy-plane.

Cross-section perpendicular to the xx-axis

See the diagram in the original worksheet below.

Schematic cross-section; dimensions vary with xx.

Original worksheet page 1: question and worked solution for 6-5-010
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Question 10 – Solution

1. Interpret the slice direction and find the bounds.

The base segments are parallel to the yy-axis, so they are vertical; integrate with respect to xx. Solving x=x2\sqrt{x}=x^2 gives x=x4x=x^4, or x(x3−1)=0x(x^3-1)=0. The real intersections occur at x=0,1x=0,1.

2. Identify width and height.

For 0≤x≤10\le x\le1, x≥x2\sqrt{x}\ge x^2. Thus each rectangle hasw(x)=x−x2,h=2.w(x)=\sqrt{x}-x^2,\qquad h=2.The height of 22 extends out of the base plane.

3. Find the area and set up the integral.

A(x)=2(x−x2),V=∫01A(x)dx=2∫01(x1/2−x2)dx.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.

∫x1/2dx=23x3/2,∫x2dx=x33.\int x^{1/2}\,dx=\frac23x^{3/2},\qquad\int x^2\,dx=\frac{x^3}{3}.V=2[23x3/2−x33]01.V=2\left[\frac23x^{3/2}-\frac{x^3}{3}\right]_0^1.

5. Apply the bounds and simplify.

V=2[(23−13)−0]=23.V=2\left[\left(\frac23-\frac13\right)-0\right]=\boxed{\frac23}.

All volumes are in cubic units.

Original worksheet page 2: question and worked solution for 6-5-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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