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Calculus IIChapter 2: Applications of IntegralsSection 2-2

Surface Area — Question 6

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Question 6

Curve: y=x2y=x^2 on 0≤x≤10\le x\le1.
Axes of rotation: first the xx-axis, then the yy-axis.
Task: Write both surface-area integrals and determine which is larger without evaluating either integral.

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Original worksheet page 1: question and worked solution for 2-2-006
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Question 6 – Solution

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Step 1: Find the common arc-length factor. y′=2x,ds=1+4x2dx.y'=2x,\qquad ds=\sqrt{1+4x^2}\,dx. Step 2: Rotate about the xx-axis. The radius is y=x2y=x^2: Sx=2π∫01x21+4x2dx.S_x=2\pi\int_0^1x^2\sqrt{1+4x^2}\,dx. Step 3: Rotate about the yy-axis. The radius is xx: Sy=2π∫01x1+4x2dx.S_y=2\pi\int_0^1x\sqrt{1+4x^2}\,dx. Step 4: Compare the integrands. For 0<x<10<x<1, x>x2.x>x^2. Multiplying by the positive common factor 2π1+4x22\pi\sqrt{1+4x^2} preserves the inequality. Therefore, Sy>Sx.\boxed{S_y>S_x}.

Original worksheet page 2: question and worked solution for 2-2-006

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