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?
 A: 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.
A: 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:
$$\int_{0}^{90}{\pi r^2(z)\,dz} = \pi\int_{0}^{90}{(\frac{45}{2}(1-\frac{z}{90}) + \frac{30}{2}\frac{z}{90})^2\,dz}$$
A: 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:
$$
V=\pi\int_{0}^{90}\left(\frac{1}{12}x+15\right)^2\,dx=\frac{64125\pi}{2}\ ft^3
$$
Wolfram Alpha's answer.
As far as I'm able to comprehend the English language, the following must be the solid whose volume you're trying to find: 

