Volume of My Valentine

Level 2

Find the volume of the half heart-shaped region bounded by the equation x 2 + ( 5 y 4 x ) 2 = 1 x^2 + (\dfrac{5y}{4} - \sqrt{x})^2 = 1 when revolved about the y axis to six decimal places.


The answer is 3.351032.

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1 solution

Rocco Dalto
Feb 18, 2018

Using the graph above the volume using the shell method is V = 2 π 4 5 0 1 x ( ( x + 1 x 2 ) ( x 1 x 2 ) ) d x = 16 π 5 0 1 x 1 x 2 d x V = 2\pi * \dfrac{4}{5} \int_{0}^{1} x((\sqrt{x} + \sqrt{1 - x^2}) - (\sqrt{x} - \sqrt{1 - x^2})) dx = \dfrac{16\pi}{5} \int_{0}^{1} x\sqrt{1 - x^2} dx

Letting x = sin ( θ ) d x = cos ( θ ) d θ V = 16 π 5 0 π 2 ( cos ( θ ) ) 2 ( sin ( θ ) ) d θ = 16 π 15 cos 3 ( θ ) 0 π 2 = 16 π 15 3.351032 x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies V = -\dfrac{16\pi}{5} \int_{0}^{\frac{\pi}{2}} (\cos(\theta))^2 (-\sin(\theta)) d\theta = \dfrac{-16\pi}{15} \cos^3(\theta)|_{0}^{\frac{\pi}{2}} = \dfrac{16\pi}{15} \approx \boxed{3.351032}

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