Given $f (x, y) = x^{2} + y^{2} - 2 x$ $f(x, y)=x^2+y^2-2x$ , how do you the volume of the solid bounded by $z = \frac{f (x, y) + f (y, x)}{2} - \frac{5}{2}, z = \pm 3 ?$ $z=(f(x, y)+f(y,x))/2-5/2, z = +-3?$

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Answer 1 · 2016-10-26T17:06:59

$9 π$ $9pi$ cubic units.

The section of this sold by a plane parallel to the xy-plane is the

circle with

center at $(\frac{1}{2}, \frac{1}{2}, z)$ $(1/2, 1/2, z)$ and radius $R (z) = \sqrt{z + \frac{3}{2}}$ $R(z)= sqrt(z+3/2)$

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.

Given f(x, y)=x^2+y^2-2xf(x,y)=x2+y2−2xf(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?z=f(x,y)+f(y,x)2−52,z=±3?z=(f(x, y)+f(y,x))/2-5/2, z = +-3?