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When solving the following problem, I could not understand why my reasoning came up with an answer that's different than the one on the solution's manual.

Question: Consider $(x,y,z)$ such that $x^2+y^2<1, x>0, 0 \le z \le 5$. This describes one half of a cylinder (a split log). Chop out a wedge out of the log along $z=2x$. Find the volume of the wedge.

My reasoning:

Wedge Volume $=$ (half the volume of the cylinder of height 2) $-$ (volume of solid of revolution found revolving the curve $x=z/2$, from 0 to 2, over the z axis).

Half Cylinder Volume: $(\pi r^2h)/2 = \pi/2 \int_0^2 (1)^2 dz = \pi$

(Volume of solid of revolution found revolving the curve x=z/2, from 0 to 2, over the z axis): $(\pi r^2h)/2 = \pi/2 \int_0^2 (z/2)^2 dz = \pi/3$

Wedge Volume: $\pi - \pi/3 = 2\pi/3$

However, the solution's manual answer is $4/3$ and it displays a different reasoning:

The slice perpendicular to the $xz$-plane are right triangles with base of length $x$ and height $z=2x$. Therefore the area of a slice is $x^2$. The volume is:

$$\int_{-1}^1 x^2 dy = \int_{-1}^1 (1-y^2) dy = 4/3.$$

While I understand the solution's manual reasoning and even find it simpler than mine, I still cannot understand why the two approaches results in different answers. What am I missing?

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Your reasoning in your first method goes wrong in that chopping a wedge isn't the same thing as taking the solid of revolution of a curve. The solid of revolution of $z = 2x$ is a cone, so if you take a cone out of the cylinder, you're not left with a wedge but a somewhat weird-looking anti-cone shape.

If we were to analyze the cross-section of the wedge that we want in cylindrical coordinates, we'd see that it grows as the angle approaches $0$, and shrinks as it approaches either $\pi/2$ or $-\pi/2$. Whereas with your solid of revolution, the cross-section stays constant at all angles.

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  • $\begingroup$ Even if I'm revolving around the z axis? My thinking was that the wedge volume would be the half the volume of revolving the rectangle of height 2 and lenght 1 around the z axis (the half cilinder) minus half of the volume of the área under the curve x=z/2 (the half cone). $\endgroup$ – MBdr Nov 18 '14 at 19:27

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