To find the volume, the function need to be rotated \(360^\circ\) on the \(y\)-axis from \(x=0\) to \(x=-50\).
Given the curve equation,
\(y=-\dfrac{1}{50}x^2\).
Volume when rotated \(360^\circ\) on the \(y\)-axis,
\(A=\int_a^b \pi x^2 \, dy\).
Express \(x^2\) in the equation of the curve,
\(\begin{aligned} y&=-\dfrac{1}{50}x^2 \\ x^2&=-50y. \end{aligned}\)
Substitute the values of \(a\), \(b\) and \(x^2\) in the formula,
\(\begin{aligned} A&=\int_{-50}^0 \pi (-50y) \, dy \\ &=-50\pi \int_{-50}^0 y \, dy \\ &=-50\pi \left[ \dfrac{y^2}{2} \right]_{-50}^0 \\ &=-50\pi \left[ 0-\dfrac{(-50)^2}{2} \right] \\ &=-50\pi [-1250] \\ &=62500\pi \text{ cm}^3. \end{aligned}\)
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