Question

If x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ, then
A. x² + y² + z² = r²
B. x² + y²− z² = r²
C. x²− y² + z² = r²
D. z² + y²− x² = r²

Answer 1

If x= rsinAcosB x²=r²sin²Acos²B

y= rsinAsinB y²=r²sin²Asin²B

z=rcosA z²=r²cos²A

x²+y²+z² = r²sin²Acos²B + r²sin²Asin²B

+r²cos²A

= r²sin²A(cos²B+sin²B)+r²cos²A

=r²sin²A+r²cos²A

=r²(sin²A+cos²A)

=r²

x²+y²+z²=r²

hence, option A is correct

Answer 2

The required "option A) $x^2+y^2+z^2 = r^2$" is correct.

Step-by-step explanation:

We have,

x = $r\sin \theta \cos \psi$                            .................(1)

y = $r \sin \theta \sin \psi$                           .................(2)

and z = $r\cos \theta$                            .................(3)

Squaring and adding (1), (2) and (3), we get

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

⇒ $x^2+y^2+z^2 = r^2\sin^2 \theta \cos^2 \psi+r^2 \sin^2 \theta \sin^2 \psi+r^2\cos^2 \theta$

⇒ $x^2+y^2+z^2 = r^2\sin^2 \theta (\cos^2 \psi+ \sin^2 \psi)+r^2\cos^2 \theta$

Using the trigonometric identity,

$\sin^2 A +\cos^2 A$ = 1

⇒ $x^2+y^2+z^2 = r^2\sin^2 \theta (1)+r^2\cos^2 \theta$

⇒ $x^2+y^2+z^2 = r^2\sin^2 \theta +r^2\cos^2 \theta$

Taking $r^2$ as common, we get

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

Again, using the trigonometric identity,

$\sin^2 A +\cos^2 A$ = 1

⇒ $x^2+y^2+z^2 = r^2(1)$

⇒ $x^2+y^2+z^2 = r^2$

Thus, the required "option A) $x^2+y^2+z^2 = r^2$" is correct.