## Application of Integration

 3.4 Application of Integration

Example
 Question

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 $$\pi$$.

 Solution (a)

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}

 Solution (b)

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}

## Application of Integration

 3.4 Application of Integration

Example
 Question

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 $$\pi$$.

 Solution (a)

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}

 Solution (b)

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}