### Video Transcript

Determine the surface area of the sphere of equation π₯ squared plus π¦ squared plus π§ squared minus 1,444 equals zero, leaving your answer in terms of π.

The surface area of a sphere is given by the expression four times π times the radius of the sphere squared. So our goal will be to use the given equation for our sphere to find its radius. To do this, weβll need to transform our equation into the so-called standard form of the equation of a sphere.

The standard form for the equation of a sphere is π₯ minus π₯ naught squared plus π¦ minus π¦ naught squared plus π§ minus π§ naught squared equals π squared. π₯ naught, π¦ naught, and π§ naught are the coordinates of the center of the sphere and π squared is the square of its radius. So from the equation of a sphere in standard form, we can directly read off π squared, which is the quantity we need for our surface area formula.

In general, putting an equation in standard form involves completing the square for each variable. However, for our equation, this is essentially already done for us. To see why, letβs expand out the term π₯ minus π₯ naught squared. π₯ minus π₯ naught squared is π₯ squared plus two π₯ naught π₯ plus π₯ naught squared. π₯ squared is quadratic in the variable π₯, two π₯ naught π₯ is linear in the variable π₯, and π₯ naught squared is a constant. Expanding out the other two terms gives us the same sort of thing. We get a quadratic term in π¦ and a quadratic term in π§, a linear term in π¦ and a linear term in π§, and a constant term from each.

Our goal is to work backwards from the expanded equation to the equation in standard form. And to do that, weβll need to find π₯ naught, π¦ naught, and π§ naught. Note that the quadratic terms donβt help us because they are unrelated to the values of π₯ naught, π¦ naught, and π§ naught. Similarly, the constant term doesnβt help because π₯ naught squared, π¦ naught squared, and π§ naught squared all act together into a single constant. However, the linear terms are exactly what we need. They donβt add across variables and they depend on the coordinates of the center of the sphere.

In particular, half the coefficient of each linear term is the corresponding coordinate for the center of the sphere. However, looking back at our original equation, we have all three quadratic terms and a single constant term but no linear terms. But the only way for linear terms not to appear in this equation is if their coefficients are zero. In other words, two π₯ naught is zero, two π¦ naught is zero, and two π§ naught is zero. So π₯ naught, π¦ naught, and π§ naught are all zero. But that means that our original equation is equivalent to π₯ minus zero squared plus π¦ minus zero squared plus π§ minus zero squared minus 1,444 equals zero.

But looking carefully, we see that the sum of these three terms is exactly the left-hand side of the equation of a sphere in standard form. So to finish putting this equation in standard form, all we need to do is add 1,444 to both sides. So π₯ minus zero squared plus π¦ minus zero squared plus π§ minus zero squared is equal to 1,444, from which then follows directly from the equation in standard form that π squared is 1,444.

So now all we do is plug this value into the surface area formula. So we have four times π times 1,444, which in terms of π is 5,776π. Itβs worth mentioning that since π₯ minus zero squared is just π₯ squared, π¦ minus zero squared is just π¦ squared, and π§ minus zero squared is just π§ squared, we couldβve recognized at the beginning that adding 1,444 to both sides of our equation takes us directly to this equation in standard form. And so 1,444 is the square of the radius of the sphere.