The volume of an object expressed in spherical coordinates is given bysin φ dr dφ dθ.The value of the integral isa)π/3b)2π/3c)π/6d)π/4Correct answer is option 'A'. Can you explain this answer?

Question

The volume of an object expressed in spherical coordinates is given bysin &phi; dr d&phi; d&theta;.The value of the integral isa)&pi;/3b)2&pi;/3c)&pi;/6d)&pi;/4Correct answer is option 'A'. Can you explain this answer?

Answer

Answer

The volume integral in spherical coordinates is given by:

V = ∫₀^2π ∫₀^π/3 ∫₀¹ r² sin(φ) dr dφ dθ.

Step 1: Integrate with respect to r

The inner integral is with respect to r from 0 to 1:

∫₀¹ r² dr = [ r³/3 ]₀¹ = 1/3.

Substituting this result, we now have:

V = ∫₀^2π ∫₀^π/3 (1/3) sin(φ) dφ dθ.

Step 2: Integrate with respect to φ

Now we integrate with respect to φ from 0 to π/3:

∫₀^π/3 sin(φ) dφ = [ -cos(φ) ]₀^π/3 = -cos(π/3) + cos(0).

Since cos(π/3) = 1/2 and cos(0) = 1, this becomes:

= -1/2 + 1 = 1/2.

Substituting this result, we get:

V = ∫₀^2π (1/3) * (1/2) dθ = (1/6) ∫₀^2π dθ.

Step 3: Integrate with respect to θ

Now we integrate with respect to θ from 0 to 2π:

∫₀^2π dθ = 2π.

So, we have:

V = (1/6) * 2π = π/3.

Conclusion:

The value of the integral is:

A: π/3.

Answer

I guess this is incorrect...as the limit of theta is always between 0 to pi ....so we cannot take its limit as 0 to pi.

Range of phi is from 0 to 2pi ...

Taking it this way ...we not give us the answer...so I believe that there's a fault in the options...

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