# Volume of a lighthouse given height and diameters

Give a lighthouse of height 90 feet, with diameters of 45 feet and 30 feet at the bottom and top respectively, what would the volume be. The lighthouse isn't a perfect cone, part of the top is cut off.

I know that the integral would be $$\int_{0}^{90} \pi r^2 dx$$

The radius would be the slope of the edge times the height on the lighthouse which would be $$45-\frac{15x}{180}$$ so the final integral is $$\int_{0}^{90} \pi \left(45-\frac{15x}{180}\right)^2 dx$$

Is my logic correct?

• Yes it is correct – BalancedTryteOperators Feb 8 at 18:56
• Note that the radius is the diameter divided by 2. – lightxbulb Feb 8 at 19:03

A much simpler approach: the shape in question is the difference of two cones with diameter and height $$(d_1, h_1) = (45, 270)$$ and $$(d_2, h_2) = (30, 180)$$. This makes the volume $$V = \frac13\pi\left(\left(\frac{45}2\right)^2270 - \left(\frac{30}2\right)^2180\right) = \frac{64125\pi}2.$$

Even if you are determined to use calculus to get the answer, at least you can do this quick calculation to check the result.

Note that the radius is equal to the diameter divided by $$2$$. In order to formalize this a bit, you can write the radius as a function of $$z$$ (a convex linear combination of the two radii): $$r(z) = \frac{45}{2}(1-\frac{z}{90}) + \frac{30}{2}\frac{z}{90}, z \in [0,90]$$ And then integrate: $$\int_{0}^{90}{\pi r^2(z)\,dz} = \pi\int_{0}^{90}{(\frac{45}{2}(1-\frac{z}{90}) + \frac{30}{2}\frac{z}{90})^2\,dz}$$

You're using the disk method to find the volume of that lighthouse. Therefore, you're going to use this formula: $$V=\pi\int_{a}^{b}[r(x)]^2\,dx$$. Your radius function is going to be $$r(x)=\frac{45/2-15}{90}x+15=\frac{1}{12}x+15$$ where $$\frac{1}{12}$$ is the slope of the line crossing the $$y$$-axis at the point $$y=15$$ and going further up, if I understand the problem correctly. Your bounds of integration are from 0 to 90:

$$V=\pi\int_{0}^{90}\left(\frac{1}{12}x+15\right)^2\,dx=\frac{64125\pi}{2}\ ft^3$$

As far as I'm able to comprehend the English language, the following must be the solid whose volume you're trying to find:

• The diameter is not the radius. – lightxbulb Feb 8 at 19:24
• oh yes, I overlooked that. Give me a second to fix that. – Michael Rybkin Feb 8 at 19:25
• Also you took the cutoff cone flipped. If he wanted to integrate from the base $0$ to some random $x$ then yours would be from the top to some $x$. – lightxbulb Feb 8 at 19:27
• I'll make a picture. – Michael Rybkin Feb 8 at 19:33
• You don't have to make a picture, you can easily check that your $r$ doesn't produce the required length at both ends of the cone, by evaluating $r(0)$ and $r(90)$. – lightxbulb Feb 8 at 19:35