A Question about a recent Problem of the Week

I was trying out the problem of the week about the polar graph r=sin⁡(4θ)sin⁡(θ)r=\frac{\sin\left(4\theta\right)}{\sin\left(\theta\right)} and its volume of revolution about the x-axis. My method is quite different from all of the solutions that have been posted, and so I'm having trouble figuring out why it doesn't work. My goal was to solve the problem without converting to rectangular coordinates. As such, I represented the area using a bunch of cones with top points at the origin and centered along the x-axis. The lateral surface area of these cones would then be integrated to obtain the volume, much in the same way that the method of cylindrical shells works. (See the diagram for a quick image -- the rotation of the triangle around the x-axis is one of the cones whose lateral surface area is being calculated and integrated.) We exclude the area of the base of each cone because the bases alone are the disks that would be used in disks/washers, and so if we included those as well we'd be covering the volume more than once.

This method has some restrictions that I figured out, but the graph in the problem shouldn't pose any difficulties for it. At least intuitively, just by visualizing the problem and the filling of the volume with the lateral surface areas of the cones, I see nothing wrong with this method.

The lateral surface area of a cone with lateral height (height from edge of the bottom to the point on the top) ll and base radius RR is πRl\pi R l. At some angle θ\theta, the lateral height of the cone on the polar graph in question is just r=sin⁡(4θ)sin⁡(θ) r=\frac{\sin\left(4\theta\right)}{\sin\left(\theta\right)} . The radius of this cone is equal to rsin⁡(θ)r \sin (\theta). Therefore the lateral surface area is πrsin⁡(θ)sin⁡(4θ)sin⁡(θ)=πsin⁡2(4θ)sin⁡(θ)\pi r \sin (\theta) \frac{\sin\left(4\theta\right)}{\sin\left(\theta\right)} =\pi \frac{\sin^2\left(4\theta\right)}{\sin\left(\theta\right)}.

Therefore, the volume of the part in the first quadrant when rotated about the x-axis is equal to π∫0π4sin⁡2(4θ)sin⁡(θ)dθ\pi\int_0^{\frac{\pi}{4}}\frac{\sin^2\left(4\theta\right)}{\sin\left(\theta\right)}d\theta. According to WolframAlpha, this has a precise value of 176π105−32π2105\frac{176\pi}{105}-\frac{32\pi\sqrt{2}}{105}. The volume of the remaining part in the second quadrant is π∫π23π4sin⁡2(4θ)sin⁡(θ)dθ\pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{\sin^2\left(4\theta\right)}{\sin\left(\theta\right)}d\theta. This has an exact value of 32π2105\frac{32\pi\sqrt{2}}{105}. The volume of the entire revolution is the sum of these two values, which is simply 176π105\frac{176\pi}{105}. But this is far smaller than the correct solution of 83π\frac 8 3 \pi.

Where did this method go wrong? I'm having trouble seeing anything, and I'm pretty sure the mechanics of the integrals are correct, so it's probably something with the method that I'm not seeing.

#Calculus

Note by Aram Lindroth
2 years, 12 months ago

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Comments

OK, I think I see what's wrong here: the infinitesimally small conical shells can't just be multiplied by dθd\theta in order to obtain the change in volume dVdV. They have a conical shape with a nonuniform thickness, so I need to take that into account. I'm working out the details now, but the result involves much more complicated integrands. that accurately represent the stranger shape of the conical shells.

Aram Lindroth - 2 years, 12 months ago

I tried this approach at first as well but I realized it wasn't working because the answer I kept getting was intuitively too small. I tried the same thing revolving the circle r=2cos(θ)r=2cos(\theta) since that has a known volume, but I couldn't get it to work.

Jeremy Galvagni - 2 years, 11 months ago
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