It is symmetry. Because of the symmetry of the solid, we can integrate half of the solid and double the volume. This allows for simpler arithmetic at the end.
Excellent question. Your cross sections would still be right triangles. Set up your integral using the same variables. The height of the triangle is parameterized by its location along the spectrum and would be unaltered. The length along the base side is what is changed now. Since the plane intersects the base circle as a line segment, the amount of the base side changed by a constant value, an amount either added to or subtracted from depending on where the plane intersects the base circle's center. It should only involve a very minor change to the integral. The area of a specific cross section should be V = 1/2*b*h*dx = 1/2*(r^2-x^2-a)^(1/2)*(r^2-x^2)^(1/2)dxx. But note that small change dramatically complicates the calculation of the antiderivative. I would recommend a numerical estimate. The original setup using isosceles triangle is a great simplifier.
For any angle of cut say theta and radius A of the tree, the Volume of wedge = (2/3)A^3 tangent theta
Yes, that is correct. For a general radius r and angle θ, the method shown here would generalize to V = 2/3*r^3*tan θ.
Hi sorry to bother but why is there a 2 outside of the integral when your first setting the integral up, thank you
It is symmetry. Because of the symmetry of the solid, we can integrate half of the solid and double the volume. This allows for simpler arithmetic at the end.
The limits -4 to +4 can be written 0 to 4 provided you multiply the integrand by 2 since you are dealing with 2 quadrants of the solid...
what if the angles slice did not pass through the midpoint of the cylinder?
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Excellent question. Your cross sections would still be right triangles. Set up your integral using the same variables. The height of the triangle is parameterized by its location along the spectrum and would be unaltered. The length along the base side is what is changed now. Since the plane intersects the base circle as a line segment, the amount of the base side changed by a constant value, an amount either added to or subtracted from depending on where the plane intersects the base circle's center. It should only involve a very minor change to the integral. The area of a specific cross section should be V = 1/2*b*h*dx = 1/2*(r^2-x^2-a)^(1/2)*(r^2-x^2)^(1/2)dxx. But note that small change dramatically complicates the calculation of the antiderivative. I would recommend a numerical estimate. The original setup using isosceles triangle is a great simplifier.
Nice explanation...
Thank you
helpful stuff!!!!!!!
I'm glad it helped.
awesome
Thanks.