Surface Area — Question 8

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Question 8

Curve: y=R2−x2y=\sqrt{R^2-x^2} on a≤x≤ba\le x\le b, where R>0R>0 and −R≤a<b≤R-R\le a<b\le R.
Axis of rotation: the xx-axis.
Task: For fixed RR, show that the surface area depends only on the zone height h=b−ah=b-a.

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Original worksheet page 1: question and worked solution for 2-2-008
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Question 8 – Solution

See the diagram in the original worksheet below.

Step 1: Differentiate. y′=−xR2−x2.y'=-\frac{x}{\sqrt{R^2-x^2}}. Step 2: Simplify the arc-length factor. 1+(y′)2=1+x2R2−x2=R2R2−x2=RR2−x2.\begin{align*} \sqrt{1+(y')^2} &=\sqrt{1+\frac{x^2}{R^2-x^2}}\\ &=\sqrt{\frac{R^2}{R^2-x^2}} =\frac{R}{\sqrt{R^2-x^2}}. \end{align*} Step 3: Calculate the surface area. S=2π∫aby1+(y′)2dx=2π∫abR2−x2RR2−x2dx=2πR∫abdx=2πR(b−a)=2πRh.\begin{align*} S&=2\pi\int_a^by\sqrt{1+(y')^2}\,dx\\ &=2\pi\int_a^b\sqrt{R^2-x^2} \frac{R}{\sqrt{R^2-x^2}}\,dx\\ &=2\pi R\int_a^bdx\\ &=2\pi R(b-a)=2\pi Rh. \end{align*} S=2πR(b−a)=2πRh\boxed{S=2\pi R(b-a)=2\pi Rh}

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

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