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Calculus IChapter 6: Application of IntegralsSection 6-5

More Volume Problems — Question 8

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Question 8

The base of a solid is the region bounded by y=2x−x2andy=0.y=\sqrt{2x-x^2} \quad\text{and}\quad y=0. Cross-sections perpendicular to the xx-axis are squares. 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 6-5-008
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Question 8 – Solution

1. Determine the bounds.

The square root is real when 2x−x2=x(2−x)≥02x-x^2=x(2-x)\ge0, giving 0≤x≤20\le x\le2. These endpoints are also the intersections with y=0y=0.

2. Find the square side length.

The cross-sections are perpendicular to the xx-axis. Their side length is the vertical segment in the base:s(x)=2x−x2−0.s(x)=\sqrt{2x-x^2}-0.

3. Find the cross-sectional area.

A(x)=s(x)2=(2x−x2)2=2x−x2.A(x)=s(x)^2=\left(\sqrt{2x-x^2}\right)^2=2x-x^2.

4. Set up and integrate the volume.

V=∫02A(x)dx=∫02(2x−x2)dx=[x2−x33]02.V=\int_0^2 A(x)\,dx=\int_0^2(2x-x^2)dx=\left[x^2-\frac{x^3}{3}\right]_0^2.

5. Evaluate and simplify.

V=(4−83)−0=12−83=43.V=\left(4-\frac83\right)-0=\frac{12-8}{3}=\boxed{\frac43}.

All volumes are in cubic units.

Original worksheet page 2: question and worked solution for 6-5-008

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

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