Vector Calculus Problem 1. Let T be the solid bounded by the plane z = 0 and the
ID: 2882296 • Letter: V
Question
Vector Calculus Problem
1. Let T be the solid bounded by the plane z = 0 and the paraboloid z = 1 x^2 y^2 .
(a) Set up a triple integral (as an iterated integral in xyz coordinates) that gives the volume of the solid T. Do NOT evaluate the integral.
(b) Using cylindrical coordinates to rewrite the triple integral in part (a). Do NOT evaluate the integral.
(c) Set up a double integral (as an iterated integral in xy coordinates) that gives the volume of the solid T. Do NOT evaluate the integral.
(d) Using polar coordinates to rewrite the double integral in part (c) and then find the volume of T.
(e) Find the surface area of the paraboloid z = 1 x^2 y^2 that lies above the plane z = 0.
Explanation / Answer
1.)
z = 0 and z = 1 x^2 y^2
1 x2 y2=0
=>x2 +y2 =1
-(1-x2) <=y<=(1-x2) , -1<= x <=1
volume = [-1 to 1][-(1-x2) to (1-x2)][0 to 1 x2 y2] dz dy dx
b)
in cylindrical coordinates
x=rcos, y=rsin
x2+y2=r2
x2+y2=1=12
0<=<=2,0<=r<=1 ,0<=z<=1-r2
dv =r dz dr d
volume =[0 to 2] [0 to 1] [0 to 1-r2] r dz dr d
c)
volume = [-1 to 1][-(1-x2) to (1-x2)] (1 x2 y2-0) dy dx
d)
volume =[0 to 2] [0 to 1](1-r2-0)r dr d
volume =[0 to 2] [0 to 1](r-r3) dr d
volume =[0 to 2][0 to 1]((1/2)r2-(1/4)r4) d
volume =[0 to 2]((1/2)12-(1/4)14)-0 d
volume =[0 to 2](1/4)d
volume =[0 to 2](1/4)
volume = (1/4)(2-0)
volume = (/2)
e)z=1 x2 y2
zx=-2x,zy=-2y
dS=[1+zx2+zy2] dA
dS=[1+4x2+4y2] dA
dS=[1+4r2] dA
surface area =[0 to 2] [0 to 1][1+4r2] r dr d
surface area =[0 to 2][0 to 1](2/3)(1/8)[1+4r2]3/2 d
surface area =[0 to 2] (1/12)[[1+4]3/2-[1+0]3/2] d
surface area =[0 to 2] (1/12)[53/2-1] d
surface area =[0 to 2] (1/12)[53/2-1]
surface area = (1/12)[53/2-1](2 -0)
surface area = (/6)[53/2-1]
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