What is the cross product of #<-3,0,1># and #<1,2,-4>#?
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"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
The cross product of (-3,0,1) and (1,2,-4) is (-2,11,-6).
We have two vectors #A=(A_x,A_y,A_z)# and #B=(B_x,B_y,B_z)#.
The cross product can be calculated nicely using matrices.
It is #det((hat(x),hat(y),hat(z)),(A_x,A_y,A_z),(B_x,B_y,B_z))=#
In the first row we have unit vectors (length 1, along each of the axes) next are coordinates of first vector, and then second vector.
#=hat(x)A_yB_z+hat(y)A_zB_x+hat(z)A_xB_y
-hat(x)A_zB_y-hat(y)A_xB_z-hat(z)A_yB_x
=hat(x)(A_yB_z-A_zB_y)+hat(y)(A_zB_x-A_xB_z)
+hat(z)(A_xB_y-A_yB_x)
=(A_yB_z-A_zB_y, A_zB_x-A_xB_z, A_xB_y-A_yB_x)#
This expression is resulting vector. The formula could be memorized by the people, who don't know matrices yet. Notice how Neither #A_x# nor #B_x# has any impact on x coordinate of the result. The same for y and z. That's the sign of perpendicularity of the result to both vectors. Notice how result disappears, when #A~~B#.
Plugging values in:
#A=(-3,0,1)# and #B=(1,2,-4)#
#A times B#
#=(0*(-4)-1*2, 1*1-(-3)(-4), (-3)*2-0*1)
=(0-2, 1-12, -6-0)=(-2,11,-6)#
