Given #f(x, y)=x^2+y^2-2x#, how do you the volume of the solid bounded by #z=(f(x, y)+f(y,x))/2-5/2, z = +-3?#

Question

Samantha Adkison · Accepted Answer

#9pi # cubic units.

The section of this sold by a plane parallel to the xy-plane is the circle with center at #(1/2, 1/2, z) #and radius #R(z)= sqrt(z+3/2)# For integration to find the volume V, choose an element in the form of a circular disc of thickness #Delta z# and radius R. The faces of this disc are parallel to the xy-plane. Now, V = limit #Delta z to 0# of #sum piR^2 Delta z=int piR^2 dz#, between the limits,from #z = -3# to z = 3.. So, #V = .pi int (z+3/2) dz#, between the limits #=pi[z^2/2+3/2z],# between #z = -3 and z = 3# #= 9pi# cubic units.

William Abdalla · Answer

undefined To find the volume of the solid bounded by $ z = \frac{(f(x, y) + f(y, x))}{2} - \frac{5}{2} $ and $ z = \pm 3 $, you would first need to determine the region over which the function $ f(x, y) $ is defined and then find the bounds of integration accordingly. After that, you would set up a triple integral over the region defined by $ z = \frac{(f(x, y) + f(y, x))}{2} - \frac{5}{2} $ and the planes $ z = \pm 3 $. Then you would integrate the function over this region with respect to $ x $, $ y $, and $ z $. This would involve evaluating the triple integral over the specified region.