- #1
real10
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x^2+y^2+z^2=4
(x+2)^2+(y-1)^2+(z+2)^2=4
find volume inside both.
thanks,
(x+2)^2+(y-1)^2+(z+2)^2=4
find volume inside both.
thanks,
The given surface is represented by the equations x^2+y^2+z^2=4 and (x+2)^2+(y-1)^2+(z+2)^2=4.
The equation represents a sphere with a radius of 2 centered at the origin for the first equation, and a sphere with a radius of 2 centered at (-2,1,-2) for the second equation.
To find the volume inside the surface, you can use the formula V = (4/3)πr^3, where r is the radius of the sphere. In this case, the radius is 2, so the volume would be (4/3)π(2)^3 = 32/3π.
The two equations represent two intersecting spheres, with the first one centered at the origin and the second one centered at (-2,1,-2). The intersection of these two spheres creates a unique shape, and finding the volume inside this shape can have practical applications in fields such as physics and engineering.
Yes, the equations can be solved for specific values of x, y, and z. For example, if we set x = 0, y = 0, and z = 2, we get a point on the surface of the first sphere. Similarly, if we set x = -2, y = 1, and z = -2, we get a point on the surface of the second sphere. This shows that the equations represent points on the respective spheres and can be solved for specific values of x, y, and z.