Find the area of one petal of the polar curve r = 5sin 2θ

Supposing that there is no direct formula available for deriving the area, we can derive the area of that region by dividing it into smaller pieces, whose area can be known directly.

Then summing all those pieces' area gives us the area of the main big region.

The polar curve r = 5sin 2θ

When r = 0, θ = 0, π/2 which corresponds to one loop.

We need the limits to provide r=0 or cos(5x) =0 so:

Therefore the area of the region enclosed by this curve is

[tex]A = \dfrac{1}{2} \int\limits^a_b {r^{2} } \, d\theta \\\\\\A = \dfrac{1}{2} \int\limits^{\pi /2}_0 {25sin^{2}2\theta } \, d\theta \\\\\\A = \dfrac{1}{2} \times 25\int\limits^{\pi /2}_0 {sin^{2}2\theta } \, d\theta \\\\\\A = \dfrac{1}{2} \times 25\int\limits^{\pi /2}_0 \dfrac {1-cos^{2}4\theta } {2}\, d\theta\\ \\\\A = \dfrac{25}{2}( \theta - \dfrac {sin4\theta} {2})\, d\theta |^{\pi /2}_0\\\\\\A = \dfrac{25}{2} \times \dfrac{\pi }{2} \\\\\\A = \dfrac{25\pi }{4}[/tex]

Thus, the area of one petal of the polar curve r = 5sin 2θ would be 25π/4.

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