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The surface area of the solid generated by the revolution of the curve r2 = a2 cos 2θ about a tangent at the pole is given by:
- a)πa2
- b)2πa2
- c)4πa2
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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The surface area of the solid generated by the revolution of the curve...
Let P(r, θ) be a point on the lemniscate
r2 = a2 cos 2θ
Let OT be a taneent at the pole
r2 = a2 cos 2θ
Let OT be a taneent at the pole
Most Upvoted Answer
The surface area of the solid generated by the revolution of the curve...
To find the surface area of the solid generated by the revolution of the curve r^2 = a^2 cos(2θ), we can use the formula for surface area of a solid of revolution:
A = 2π∫[α,β] (r(θ) * √(r(θ)^2 + (dr(θ)/dθ)^2)) dθ
In this case, we have r(θ) = √(a^2 cos(2θ)), and dr(θ)/dθ = -2a^2 sin(2θ).
Substituting these values into the formula, we get:
A = 2π∫[α,β] (√(a^2 cos(2θ)) * √((√(a^2 cos(2θ)))^2 + ((-2a^2 sin(2θ))^2))) dθ
Simplifying, we have:
A = 2π∫[α,β] (√(a^2 cos(2θ)) * √(a^2 cos(2θ) + 4a^4 sin^2(2θ))) dθ
A = 2π∫[α,β] (√(a^4 cos^2(2θ) + 4a^4 sin^2(2θ))) dθ
A = 2π∫[α,β] (√(a^4 (cos^2(2θ) + 4sin^2(2θ)))) dθ
A = 2π∫[α,β] (a^2√(cos^2(2θ) + 4sin^2(2θ))) dθ
Now, we can use a trigonometric identity to simplify the integrand:
cos^2(2θ) + 4sin^2(2θ) = 1 + 3sin^2(2θ)
A = 2π∫[α,β] (a^2√(1 + 3sin^2(2θ))) dθ
This integral can be quite difficult to evaluate analytically, so it may be necessary to use numerical methods or approximations to find the exact surface area.
A = 2π∫[α,β] (r(θ) * √(r(θ)^2 + (dr(θ)/dθ)^2)) dθ
In this case, we have r(θ) = √(a^2 cos(2θ)), and dr(θ)/dθ = -2a^2 sin(2θ).
Substituting these values into the formula, we get:
A = 2π∫[α,β] (√(a^2 cos(2θ)) * √((√(a^2 cos(2θ)))^2 + ((-2a^2 sin(2θ))^2))) dθ
Simplifying, we have:
A = 2π∫[α,β] (√(a^2 cos(2θ)) * √(a^2 cos(2θ) + 4a^4 sin^2(2θ))) dθ
A = 2π∫[α,β] (√(a^4 cos^2(2θ) + 4a^4 sin^2(2θ))) dθ
A = 2π∫[α,β] (√(a^4 (cos^2(2θ) + 4sin^2(2θ)))) dθ
A = 2π∫[α,β] (a^2√(cos^2(2θ) + 4sin^2(2θ))) dθ
Now, we can use a trigonometric identity to simplify the integrand:
cos^2(2θ) + 4sin^2(2θ) = 1 + 3sin^2(2θ)
A = 2π∫[α,β] (a^2√(1 + 3sin^2(2θ))) dθ
This integral can be quite difficult to evaluate analytically, so it may be necessary to use numerical methods or approximations to find the exact surface area.
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