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Worked solution

Transform a square-root exponential volume

Sketch y=exy=e^{\sqrt x}, x≥0x\ge0. mark Rotate the region bounded by this curve, x=4x=4 and both axes completely about the x-axis. Use t=xt=\sqrt x to show the volume is 2π∫02te2t dt2\pi\int_0^2te^{2t}\,dt, and evaluate it exactly. marks

[2014/YJC/II/3(c)]

CJC · Review problems · Question 11

Answer

V=π2(3e4+1)V=\tfrac\pi2(3e^4+1).

Sketch the increasing curve and the bounded region

The curve begins at (0,1)(0,1), increases, and passes through (4,e2)(4,e^2). It has a vertical one-sided tangent at x=0x=0. Its concavity changes at x=1x=1; plotting the actual formula retains the early concave-down segment as well as the later rise. The requested region is underneath the curve for 0≤x≤40\le x\le4.

How did this part go?

Change the radius, differential and limits

V=π∫04e2xdxV=\pi\int_0^4e^{2\sqrt x}dx. With x=t2x=t^2, dx=2t dtdx=2t\,dt, x=0↦t=0x=0\mapsto t=0, x=4↦t=2x=4\mapsto t=2, this is 2π∫02te2tdt2\pi\int_0^2te^{2t}dt.

How did this part go?

Evaluate by parts

∫te2tdt=12te2t−14e2t+C\int te^{2t}dt=\tfrac12te^{2t}-\tfrac14e^{2t}+C. Therefore V=2π[e4−(e4−1)/4]=π2(3e4+1)V=2\pi[e^4-(e^4-1)/4]=\tfrac\pi2(3e^4+1) cubic units.

How did this part go?

Compare your sketch

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Increasing curve exp(sqrt(x)) begins at (0,1), passes through (4,e²), and has a vertical tangent at zero. The region under the curve from x=0 to 4 is shaded.
Increasing curve exp(sqrt(x)) begins at (0,1), passes through (4,e²), and has a vertical tangent at zero. The region under the curve from x=0 to 4 is shaded.
Increasing curve exp(sqrt(x)) begins at (0,1), passes through (4,e²), and has a vertical tangent at zero. The region under the curve from x=0 to 4 is shaded.

Common mistakes

The substitution changes the bounds from 0,4 to 0,2 and introduces the factor 2t.

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