Application of Integration

3.4 Application of Integration
 
This image illustrates the applications of integration. It features a central box labeled 'Application of Integration' with three arrows pointing to different examples: 'Designing Containers' with an image of stacked containers, 'Designing Domes and Arches' with an image of a dome structure, and 'Rainfall Accumulation' with an image of rain clouds. The Pandai logo is also present at the top left.
 
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. 

Visual representation of cross-section of a rattan food cover which is parabolic in shape with height of 50 cm and diameter of 100 cm.

(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,

Graph depicting a curved line representing the equation y=-kx, with labeled endpoints and parameters.

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
 
This image illustrates the applications of integration. It features a central box labeled 'Application of Integration' with three arrows pointing to different examples: 'Designing Containers' with an image of stacked containers, 'Designing Domes and Arches' with an image of a dome structure, and 'Rainfall Accumulation' with an image of rain clouds. The Pandai logo is also present at the top left.
 
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. 

Visual representation of cross-section of a rattan food cover which is parabolic in shape with height of 50 cm and diameter of 100 cm.

(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,

Graph depicting a curved line representing the equation y=-kx, with labeled endpoints and parameters.

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,

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