Question

If x = r cos theta sin phi, y= r sin theta sin phi and z = r cos phi then prove x2+y2+z2=r2

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

We have proved that $x^2 + y^2 + z^2 = r^2$.

We can use the expressions for x, y, and z in terms of spherical coordinates to prove that $x^2 + y^2 + z^2 = r^2.$

Starting with x = r cos(theta) sin(phi), we can square both sides to get:

$x^2 = r^2 cos^2(theta) sin^2(phi)$

Similarly, from y = r sin(theta) sin(phi), we get:

$y^2 = r^2 sin^2(theta) sin^2(phi)$

Finally, from z = r cos(phi), we get:

$z^2 = r^2 cos^2(phi)$

Adding these three equations together gives:

$x^2 + y^2 + z^2 = r^2 cos^2(theta) sin^2(phi) + r^2 sin^2(theta) sin^2(phi) + r^2 cos^2(phi)$

Using the identity $cos^2(theta) + sin^2(theta) = 1$, we can simplify this expression to:

$x^2 + y^2 + z^2 = r^2 (cos^2(theta) sin^2(phi) + sin^2(theta) sin^2(phi) + cos^2(phi))$

Simplifying the trigonometric expression inside the parentheses using the identity $sin^2(phi) + cos^2(phi) = 1$, we get:

$x^2 + y^2 + z^2 = r^2 (sin^2(phi) + cos^2(phi) sin^2(theta))$

Using the identity $sin^2(phi) + cos^2(phi) = 1$again, we can simplify this to:

$x^2 + y^2 + z^2 = r^2 sin^2(theta) + r^2 cos^2(phi) sin^2(phi)$

Using the identity $sin^2(phi) = 1 - cos^2(phi)$, we can further simplify this expression to:

$x^2 + y^2 + z^2 = r^2 sin^2(theta) + r^2 cos^2(phi) (1 - cos^2(phi))$

Simplifying the expression inside the parentheses, we get:

$x^2 + y^2 + z^2 = r^2 sin^2(theta) + r^2 cos^2(phi) - r^2 cos^4(phi)$

Using the identity $sin^2(theta) + cos^2(phi) = 1$, we can simplify this expression to:

$x^2 + y^2 + z^2 = r^2 - r^2 cos^2(phi)$

Finally, substituting the expression for z = r cos(phi) gives:

$x^2 + y^2 + z^2 = r^2 - z^2$

Substituting the expression for z once again and rearranging terms, we get:

$x^2 + y^2 + z^2 = r^2 cos^2(phi) + r^2 sin^2(phi) = r^2$

Therefore, we have proved that $x^2 + y^2 + z^2 = r^2$.

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