Porisma

Worked solution

Subtract the inner solid from the outer solid

Find the value of the volume of the solid generated when the region enclosed between y=x2y=x^2 and y=xy=\sqrt x is rotated about the x-axis through 2π2\pi radians.

CJC · Lecture · Example 17

Find and check the intersections

With x≥0x\ge0, solve x2=xx^2=\sqrt x. Squaring gives x4=xx^4=x, so x(x3−1)=0x(x^3-1)=0 and x=0x=0 or x=1x=1. Both satisfy the original equation. Between them, x≥x2≥0\sqrt x\ge x^2\ge0.

How did this part go?

Use outer squared minus inner squared

For a full turn, outer radius R=xR=\sqrt x and inner radius r=x2r=x^2. Hence V=π∫01[(x)2−(x2)2] dxV=\pi\int_0^1[(\sqrt x)^2-(x^2)^2]\,dx. This is the larger solid volume minus the inner solid volume, as the Teacher annotates; it is not π∫01(x−x2)2 dx\pi\int_0^1(\sqrt x-x^2)^2\,dx.

How did this part go?

Evaluate and round

The source gives 0.9420.942 cubic units to three significant figures. Independently, V=π∫01(x−x4) dx=π[x2/2−x5/5]01=3π/10≈0.942477796V=\pi\int_0^1(x-x^4)\,dx=\pi[x^2/2-x^5/5]_0^1=3\pi/10\approx0.942477796, agreeing with the source answer.

How did this part go?

Compare your sketch

Expand diagram

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On [0,1], y=sqrt(x) lies above y=x squared. The shaded region rotates about the x-axis; its washer has outer radius sqrt(x) and inner radius x squared, giving volume 3π/10.
On [0,1], y=sqrt(x) lies above y=x squared. The shaded region rotates about the x-axis; its washer has outer radius sqrt(x) and inner radius x squared, giving volume 3π/10.
On [0,1], y=sqrt(x) lies above y=x squared. The shaded region rotates about the x-axis; its washer has outer radius sqrt(x) and inner radius x squared, giving volume 3π/10.

Common mistakes

Use R squared minus r squared, not the square of R minus r. Keep the square root in the original upper curve before simplifying its squared radius.

Read the hint