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student93
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Homework Statement
See the problem attached in this post.
Homework Equations
See the problem attached in this post.
The Attempt at a Solution
I set my limits of integration with respect to z axis and got an upper limit of 2 since that's the vertex point/height of the pyramid and my lower limit as 0 since the lowest possible point in regards to the z axis is 0. The area of a square is s^2 and I set my integrand as ∫s^2 dz, from 0 to 2 and got 8/3, which is actually the correct answer. However, this is just a coincidence since I just realized you can't take the integral of s^2 with respect to the z axis (It's necessary to convert s into some term of z). How exactly do I convert s into a term of z so that I can set up the correct integrand?
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