Can someone show me how to do this triple integral question please? Thank you! B
ID: 2878587 • Letter: C
Question
Can someone show me how to do this triple integral question please? Thank you!
By changing to spherical coordinates, the value of integral integral integral_D e^(x^2 + y^2 + z^3)^3/2/2 dV, where D is the region in the first octant defined by the inequalities x^2 + y^2 + z^2 greaterthanorequalto 1, x^2 + y^2 + z^2 lessthanorequalto 3, z^2 greaterthanorequalto 3(x^2 + y^2), and 2x lessthanorequalto y lessthanorequalto 4x is (A) 2.61169 (B) 2.17085 (C) 1.59237 (D) 2.58251 (E) 0.23025 (F) 2.89263 (G) 2.45783 (H) 1.14716Explanation / Answer
D is region in in first octant
in spherical coordinates
x=sincos,y=sinsin,z=cos
x2+y2+z2=2
12=1<=x2+y2+z2<=3=32
==>1<=<=3
z2>=3(x2+y2)
2cos2>=32sin2
tan2<=1/3
tan<=1/3
<=/6
2x<=y<=4x
2<=y/x<=4
2<=tan()<=4
tan-1(2)<=<=tan-1(4)
dv=2sin d d d
0<=<=2,0<=<=/6,1<=<=3
D e((x^2 +y^2 +z^2)^(3/2))/2 dv
= [tan-1(2) to tan-1(4)] [0 to /6] [1 to 3] e((^2)^(3/2))/22sin d d d
= [tan-1(2) to tan-1(4)] [0 to /6] [1 to 3] e(^3)/22sin d d d
= [tan-1(2) to tan-1(4)] [0 to /6][1 to 3] (2/3)e(^3)/2sin d d
= [tan-1(2) to tan-1(4)] [0 to /6] (2/3)(e(33)/2-e(1^3)/2)sin d d
= [tan-1(2) to tan-1(4)][0 to /6] (2/3)(e(33)/2-e1/2)(-cos) d
= [tan-1(2) to tan-1(4)] (2/3)(e(33)/2-e1/2)(-cos(/6) +cos0) d
= [tan-1(2) to tan-1(4)] (2/3)(e(33)/2-e1/2)(-(3 /2) +1) d
= [tan-1(2) to tan-1(4)] (2/3)(e(33)/2-e1/2)(-(3 /2) +1)
= (2/3)(e(33)/2-e1/2)(-(3 /2) +1)(tan-1(4)-tan-1(2))
=0.23025