calc III :sphere problem, completing the square?

Question

I've been having a lot of difficulty with this 3 dimensions. Can you please help!!
Show the equation 4x^2+4y^2+4z^2-12x+3y-5z =0 represents a sphere and determine its center and radius.

Also explain why the equation x^2+y^2-2xy +5 = 0 does not represent a sphere

Lastly, Describe the region represented by the equation x^2 +y^2+z^2 - 4x +2y - 6z + a > 0 for the following values of a:

(j) a = 0 (ii) a = 14 (i) a = 20.

any help is greatly appreciated and thanks alot for your time

Answer 1

1) Show the equation 4x² + 4y² + 4z² - 12x + 3y - 5z = 0
represents a sphere and determine its center and radius.

This works like a circle except in three dimensions. Complete the square in each variable.

4x² + 4y² + 4z² - 12x + 3y - 5z = 0
(4x² - 12x) + (4y² + 3y) + (4z² - 5z) = 0

Divide thru by 4.

(x² - 3x) + [y² + (3/4)y] + [z² - (5/4)z] = 0

(x² - 3x + 9/4) + [y² + (3/4)y + 9/64] + [z² - (5/4)z + 25/64]
= 9/4 + 9/64 + 25/64

(x - 3/2)² + (y + 3/8)² + (z - 5/8)² = 178/64

The center of the sphere is (h, k, p) = (3/2, -3/8, 5/8) and the radius is √178 / 8.
______________

2) Explain why the equation x² + y² - 2xy + 5 = 0 does not represent a sphere.

The value of z is unrestricted. So the equation would be some kind of cylindrical shape.
________________

3) Lastly, Describe the region represented by the equation
x² + y² + z² - 4x + 2y - 6z + a > 0 for the following values of a:

(i) a = 0 (ii) a = 14 (iii) a = 20.

x² + y² + z² - 4x + 2y - 6z + a > 0
x² + y² + z² - 4x + 2y - 6z > -a
(x² - 4x) + (y² + 2y) + (z² - 6z) > -a
(x² - 4x + 4) + (y² + 2y + 1) + (z² - 6z + 9) > -a + 4 + 1 + 9
(x - 2)² + (y + 1)² + (z - 3)² > -a + 14

(i) a = 0
-a + 14 = 0 + 14 = 14
Sphere with center (2, -1, 3) and radius √14.

(ii) a = 14
-a + 14 = -14 + 14 = 0
The single point (2, -1, 3).

(iii) a = 20
-a + 14 = -20 + 14 = -6
There are no real solutions.

Answer 2

Completing The Square Calc

Answer 3

utilizing x^2 + y^2 + z^2 = one million, it suffices to double the area of the better heiemisphere z = ?(one million - x^2 - y^2). So, the floor area equals 2 * ?? ?[one million + (z_x)^2 + (z_y)^2] dA = 2 ?? ?[one million + (-x/?(one million - x^2 - y^2))^2 + (-y?(one million - x^2 - y^2))^2] dA = 2 ?? ?[one million + (x^2 + y^2)/(one million - x^2 - y^2)] dA = 2 ?? ?[((one million - x^2 - y^2) + (x^2 + y^2))/(one million - x^2 - y^2)] dA = 2 ?? dA / ?(one million - x^2 - y^2). Now, convert to polar coordinates. because of the fact the region of integration is the unit disk, r = 0 to one million. ==> 2 ?(t = 0 to 2?) ?(r = 0 to one million) (r dr dt) / ?(one million - r^2) = 2 * 2? * -?(one million - r^2) {for r = 0 to one million} = 4?. i wish this permits!