Consider the solid in xyz-space, which contains all points (x, y, z) whose z-coordinate satisfies 0≤z≤4−x2−y2 Which statements do hold? a) The solid is a sphere b) The solid is apyramid c) Its volume is 8π d) Its volume is 16π3

Question

Consider the solid in xyz-space, which contains all points (x, y, z) whose z-coordinate satisfies  0≤z≤4−x2−y2  Which statements do hold?  a) The solid is a sphere  b) The solid is apyramid  c) Its volume is 8π  d) Its volume is 16π3

Carl Swisher · Accepted Answer

Step 1 ∫∫∫dv ⇒∫∫∫dzdydx z=0 to z=4−x2−y2 ⇒∫∫(∫04−x2−y2dz)dydx ⇒∫∫(4−x2−y2)dydx Change in folar co-ordinates, tre get x=rcos⁡θ y=rsin⁡θ r=0 to r=2 dxdy=rdrdθ θ=0 to θ=2π ⇒∫02π∫02(4−r2)rdrdθ Let 4−r2=t⇒rdr=−dt2 t=4 to t=4−4=0 ⇒∫02π(∫40−tdt2)dθ⇒12∫02π(∫04tdt)dθ ⇒12∫02π(t22)04dθ⇒12∫02π8dθ ⇒4∫02πdθ⇒4(2π) ⇒2π

Carl Swisher · Accepted Answer

Here, take limit: 0≤4−x2−y2 x2+y2≤4 And we know, (4−x2−y2)max=4 when x=0 and y=0 So, 0≤z≤4 For each value of z≥0, it will be a disk of at each value of z∈R. Now, volume in =πBr2h =πB×(2)2×(4)=16πB

nick1337 · Accepted Answer

Step 1 Consider the solid  in xyz-space which contains all points (x,y,z) whose z-coordinate satisfies 0≤z≤4−x2−y2 Step 2 The solid is paraboloid. We use cylindrical coordinates  x=rcos⁡t,y=rsin⁡t,0≤z≤4−r2,0≤t≤2π and z=0 gives 4−r2=0 ⇒r2=4 ⇒r=2 ⇒0≤r≤2 The volume of the solid is given by V=∫∫∫SdV=∫02∫02π∫04−r2rdzdtdr=∫02∫02π|z|04−r2rdtdr=∫02∫02π(4−r2)rdtdr=∫02r(4−r2)[t]02πdr=∫02r(4−r2)[2π−0]dr=2π∫02(4r−r3)dr=2π[2r2−r44]02=2π[2×22−244−0]=2π[8−4]=8π Thus, volume of the solid is 8π. Hence, option (c) is correct.

Consider the solid in xyz-space, which contains all points (x,

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