### Video Transcript

Find the Laplacian of the function π of π₯, π¦, π§ equals π to the negative π₯ squared minus π¦ squared minus π§ squared in terms of the spherical coordinates π, π, and π.

There are two approaches we can take for this problem. The first approach is to take the Laplacian in Cartesian coordinates and then convert to spherical coordinates. Letβs try this method first. Recall that the Laplacian of a function β squared π is equal to the divergence of the gradient of π, which in Cartesian coordinates is given by β two π by βπ₯ squared plus β two π by βπ¦ squared plus β two π by βπ§ squared. So to begin with, partial π by partial π₯ is partial by partial π₯ of π to the negative π₯ squared minus π¦ squared minus π§ squared. So we differentiate with respect to π₯ and treat π¦ and π§ as a constant, giving negative two π₯ times π to the negative π₯ squared minus π¦ squared minus π§ squared.

Therefore, β two π by βπ₯ squared is given by taking the partial derivative with respect to π₯ again of negative two π₯π to the negative π₯ squared minus π¦ squared minus π§ squared. Differentiating partially with respect to π₯ and using the product rule gives us negative two times π to the negative π₯ squared minus π¦ squared minus π§ squared plus four π₯ squared π to the negative π₯ squared minus π¦ squared minus π§ squared.

We can take out a common factor of two times the exponent to give two π to the negative π₯ squared minus π¦ squared minus π§ squared times two π₯ squared minus one. Now, we could repeat this process for β two π by βπ¦ squared and β two π by βπ§ squared. But we can save some time by recognizing the symmetry of the function. We can see that the function π has the same dependence on π¦ and π§ as it does on π₯. So differentiating partially with respect to π¦ and π§ will have the same result but with π¦ and π§ replacing π₯, respectively.

So our expressions for β two π by βπ¦ squared and β two π by βπ§ squared will be the same as the expression for β two π by βπ₯ squared, but with π¦ and π§ replacing π₯. Adding these all together and taking out the exponent as a common factor gives us the Laplacian of π equal to two π to the negative π₯ squared minus π¦ squared minus π§ squared times two π₯ squared plus two π¦ squared plus two π§ squared minus three. Letβs take out a common factor of negative one from the exponent and a common factor of two from the π₯ squared plus π¦ squared plus π§ squared inside the parentheses.

In spherical coordinates, the radius π squared is given by π₯ squared plus π¦ squared plus π§ squared. So these two terms here are, in fact, just equal to π squared. So this gives us our answer. The Laplacian of π, β squared π, is equal to two π to the negative π squared times two π squared minus three. An alternative method is to first convert to spherical coordinates and then take the Laplacian. As mentioned before, negative π₯ squared minus π¦ squared minus π§ squared is just negative π squared. So in spherical coordinates, the function π of π, π, π becomes simply π to the negative π squared.

This greatly simplifies the function π. But the difficulty with this method is that you need to either remember or be able to derive the Laplacian in spherical coordinates. In spherical coordinates, the Laplacian of π, β squared π, is given by one over π squared partial by partial π of π squared partial π by partial π plus one over π squared sin π partial by partial π of sin π partial π by partial π plus one over π squared sin squared π β two π by βπ squared. The good news is that in this case, this will greatly simplify taking the Laplacian of π since π depends only on π. So any partial terms with respect to either π or π will be zero.

This means that we can disregard both the second and third term, as they will both be equal to zero. First, we will need to find partial π by partial π, which is given by negative two π times π to the negative π squared. So then the Laplacian of π is given by one over π squared times partial by partial π of π squared times negative two ππ to the negative π squared. This simplifies to one over π squared times partial by partial π of negative two π cubed π to the negative π squared.

Differentiating with respect to π, we get one over π squared times negative six π squared π to the negative π squared plus four π to the fourth times π to the negative π squared. Multiplying out the parentheses and taking out a factor of π to the negative π squared, but not a factor of two this time, gives us π to the negative π squared times four π squared minus six, as before.