consider the cylinder above the circle (x-1/2)^2 + y^2=1/4 in the xy plane and below z=x+1. Compute its volume

Question

To compute its volume using polar coordinates system.

Answer 1

its (1/4)pi cubic units

Answer 2

So dV =dAdz the z limits: the xy-airplane and the airplane 2z = 4 + x, it is z=0 to z=2 + x/2. V = ?(2 + x/2) dA after the z integration In polar coordinates dA = rdrd? and utilising x = rcos?, y rsin?, x^2+y^2 = 2x turns into r^2 = 2r cos? or r = 2cos? V = ??(2 + (rcos?)/2) rdrd? The r limits are from r = 0 to r = 2cos? because of the fact the cylinder is predicated at (a million,0), and the x=0 airplane is tangential to the cylinder, the obstacles of ? are -?/2 to ?/2 { word: x^2+y^2 = 2x ==> (x-a million)^2 + y^2 = a million, centre at (a million,0), radius a million} V = ??[r = 0 to r = 2cos?](2r + (r^2 cos?)/2) drd? ...= ?[r = 0 to r = 2cos?](r^2 + (r^3 (cos?)/6) d? ...= ?[? = -?/2 to ?/2]{(2cos?)^2 + (2cos?)^3 (cos?)/6)} d? ...= 4?[? = -?/2 to ?/2]{(cos?)^2 + (cos?)^4/3} d? cos 2? = (cos?)^2 - (sin?)^2 = 2 (cos?)^2 - a million or (cos?)^2 = (cos 2? + a million)/2 (cos?)^4 = (cos 2? + a million)^2 /4 = a million/4 {(cos2?)^2 + 2cos2? + a million} ............= a million/4 {(cos 4? + a million)/2 + 2cos2? + a million} the place we used (cos2?)^2 = (cos 4? + a million)/2 So (cos?)^2 + (cos?)^4/3 = (cos 2? + a million)/2 + a million/4 {(cos 4? + a million)/2 + 2cos2? + a million}/3 .....................................= (cos 4?)/24 + 2/3 cos 2? + a million/2 + a million/24 + a million/12 and V = 4?[? = -?/2 to ?/2]{(cos?)^2 + (cos?)^4/3} d? turns into ...= 4?[? = -?/2 to ?/2]{(cos 4?)/24 + 2/3 cos 2? + 5/8} d? ?[? = -?/2 to ?/2](cos 4?) d? = (sin4?)/4 = 0 for the given limits and in addition ?[? = -?/2 to ?/2](cos 2?) d? = (sin2?)/2 = 0 for the given limits. So V = 4?[? = -?/2 to ?/2]{5/8} d? and V = 4*5/8 * ? = 5?/2 <== Ans wish this helps.

Answer 3

The volume of a cylinder is πr²h.

V = πr²h = π(1/2)²(1/2 + 1) = π(1/4)(3/2) = 3π/8

The average height of the sphere is 3/8. If you cut off the top of the cylinder at a height of 3/8, the excess to the right of the line x = 1/2, when you flip it over, will exactly fill the deficiency to the left of the line, leaving you with a cylinder with constant height 3/2.