Three volume of revolution with shell method problems

[clairez93]

Homework Statement

1. Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.

$$y= \frac{1}{x}, x = 1, x = 2, y = 0$$


2. A solid is generated by revolving the region bounded by $$y = \frac{1}{2}x^{2}$$ and $$y = 0$$ about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-fourth of the volume is removed. Find the diameter of the hole.

3. A hole is cut through the center of a sphere of radius r. The height of the remaining spherical ring is h, as shown in the figure (see attachment). Show that the volume of the ring is $$V = \frac{\pi h^{3}}{6}$$. (Note: The volume is independent of r)

The Attempt at a Solution

1. $$V = 2\pi\int^{1/2}_{0}y dy = \frac{\pi}{4}$$

Book Answer says:

$$V = 2\pi\int^{1/2}_{0}y dy + \int^{1}_{1/2} y(\frac{1}{y} - 1) dy] = \frac{\pi}{2}$$

I don't know where the second integral came from.

2. This one seems very tricky. I tried to find the volume of the the solid like this:

$$V = 2\pi\int^{2}_{0} |x(2-\frac{1}{2}x^{2})| dx = 4$$

Then I multiplied that by 1/4 to get 1.

Then I wasn't sure how to set up an expression to solve for the diameter after this.

3. I have absolutely no idea how to set this one up.

 

[zcd]

First of all, the equation for shell method is $$2\pi\int_{a}^{b}p(x)h(x)\,dx$$ for p(x) = radius and h(x) = height. The interval [a,b] is practically the interval of the radius you integrate over.

Shell method is done parallel to the revolving axis. Since the axis of rotation is horizontal line through y=0 (the x axis), you must change equations into x=g(y) form.

The need for two integrals becomes apparent once you make a sketch of the graph. Using shell method, you have to separate the integral into a curved wedge section on interval y=[0,1/2] and a rectangular section on interval y=[1/2,1].
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For #2, you already have most of it. By dividing by 4, you've solved for the "volume" of the hole. Recall that shell method is integral over an interval that is rotated, so just set up the integral as $$V(a)-V(0) = 2\pi\int^{a}_{0}|x(2-\frac{1}{2}x^{2})|dx=1$$ by fundamental theorem of calculus and solve for a radius a.
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For #3, start by drawing a diagram relating sphere with radius r to the height h of its hole (of radius a). You'll soon discover by pythagorean theorem that $$r^{2}=\frac{h^{2}}{4}+a^{2}$$.

Next, note that the volume of a sphere of radius r given by shell method is $$V=2(2\pi\int_{0}^{r}x\sqrt{r^{2}-x^{2}})\,dx=4\pi\int_{0}^{r}x\sqrt{r^{2}-x^{2}}\,dx$$.

Using experience from part 2, the volume of a sphere (of radius r) with a hole (of radius a) bored into it will be $$V=4\pi\int_{a}^{r}x\sqrt{r^{2}-x^{2}}\,dx$$.

After the integration, recall the relation between r, h, and a where $$a^{2}=r^{2}-\frac{h^{2}}{4}$$.

 

[clairez93]

Thank you, I have a very loose understanding for the visualization of #2 and #3. Although I somewhat see it, I don't fully grasp the idea that the shells are of the same shell height and radius for the solids with the hole and the solids without the hole. I was under the impression that the shell used to find the volume of the hole would be different from that used for the whole solid.

Could you by any chance, make this idea more clearer to me?

 

[zcd]

There's no reason for the formula to be different. Take #3 for example: imagine that you have a sphere of some material. If you drill a hole into that sphere, then what you take out will have a definite volume. Since shell method is summing up shells that are parallel to the axis of rotation, you can use the same formula to measure what's drilled out of the sphere as long as the stuff removed is made of shells parallel to the axis of rotation (just like the sphere itself).