The diagram below shows the cross-section of a rattan food cover which is parabolic in shape and is presented by the equation \(y=-kx^2\), where \(y\) is the height, in centimetres, and \(x\) is the radius of the food cover, in centimetres. 

(a) Show that \(k=\dfrac{1}{50}\).
(b) Find the internal volume of the food cover, in terms of \(\pi\).

Given the equation of the curve \(y=-kx^2\),

Since there is no constant variable in the equation, then the curve must only intersect with the origin \(O(0,0)\).

Sketch the function graph,

To find value of \(k\), substitute one point on the curve to the equation of the curve,

\(\begin{aligned} y&=-kx^2 \\ -50&=-k(50)^2 \\ -50&=-2500k \\ k&=\dfrac{50}{2500} \\ k&=\dfrac{1}{50}. \end{aligned}\)

​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,

\(\begin{aligned} A&=\int_{-50}^0 \pi (-50y) \, dy \\ &=-50\pi \int_{-50}^0 y \, dy \\ &=-50\pi \left[ \dfrac{y^2}{2} \right]_{-50}^0 \\ &=-50\pi \left[ 0-\dfrac{(-50)^2}{2} \right] \\ &=-50\pi [-1250] \\ &=62500\pi \text{ cm}^3. \end{aligned}\)

The diagram below shows the cross-section of a rattan food cover which is parabolic in shape and is presented by the equation \(y=-kx^2\), where \(y\) is the height, in centimetres, and \(x\) is the radius of the food cover, in centimetres. 

(a) Show that \(k=\dfrac{1}{50}\).
(b) Find the internal volume of the food cover, in terms of \(\pi\).

Given the equation of the curve \(y=-kx^2\),

Since there is no constant variable in the equation, then the curve must only intersect with the origin \(O(0,0)\).

Sketch the function graph,

To find value of \(k\), substitute one point on the curve to the equation of the curve,

\(\begin{aligned} y&=-kx^2 \\ -50&=-k(50)^2 \\ -50&=-2500k \\ k&=\dfrac{50}{2500} \\ k&=\dfrac{1}{50}. \end{aligned}\)

​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,

\(\begin{aligned} A&=\int_{-50}^0 \pi (-50y) \, dy \\ &=-50\pi \int_{-50}^0 y \, dy \\ &=-50\pi \left[ \dfrac{y^2}{2} \right]_{-50}^0 \\ &=-50\pi \left[ 0-\dfrac{(-50)^2}{2} \right] \\ &=-50\pi [-1250] \\ &=62500\pi \text{ cm}^3. \end{aligned}\)