Prove this integral?

Question

Prove that
∫(-r,r) 2√(r² - x²) dx = πr²

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Answer 1

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∫(-r,r) 2√(r² - x²) dx = πr²
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Let x = rsinθ thus dx = rcosθ dθ
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When x = -r, sinθ = -1, θ = -π/2
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When x = r, sinθ = 1, θ = π/2
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So ∫(-r,r) 2√(r² - x²) dx
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= ∫(-π/2,π/2) 2√(r² - r²sin²θ) rcosθ dθ
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= ∫(-π/2,π/2) 2r²√(1 - sin²θ) cosθ dθ Now sin²θ + cos²θ = 1
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= r²∫(-π/2,π/2) 2cosθ cosθ dθ as 1 - sin²θ = cos²θ
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= r²∫(-π/2,π/2) 2cos²θ dθ Now cos 2θ = 2cos²θ - 1
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= r²∫(-π/2,π/2) (cos2θ + 1)dθ as 2cos²θ = cos 2θ + 1
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= r² [ ½sin2θ + θ] (-π/2,π/2)
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= r² [(½sinπ + π/2) - (½sin(-π) - π/2)]
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= r² [(0 + π/2) - (0 - π/2)]
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= πr²
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QED!!!
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Hopes that this helps!! ^^!!!
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Answer 2

∫(-r,r) 2√(r² - x²) dx

Let x = rsinθ thus dx = rcosθ . dθ
When x = -r, sinθ = -1, θ = -π/2
When x = r, sinθ = 1, θ = π/2

So ∫(-r,r) 2√(r² - x²) dx
= ∫(-π/2,π/2) 2√(r² - r²sin²θ) rcosθ dθ
= ∫(-π/2,π/2) 2r²√(1 - sin²θ) cosθ dθ Now sin²θ + cos²θ = 1
= r²∫(-π/2,π/2) 2cosθ cosθ dθ as 1 - sin²θ = cos²θ
= r²∫(-π/2,π/2) 2cos²θ dθ Now cos 2θ = 2cos²θ - 1
= r²∫(-π/2,π/2) (cos2θ + 1)dθ as 2cos²θ = cos 2θ + 1
= r² [ ½sin2θ + θ] (-π/2,π/2)
= r² [(½sinπ + π/2) - (½sin(-π) - π/2)]
= r² [(0 + π/2) - (0 - π/2)]
= πr²