Area Mania #2

Calculus Level 3

Find the enclosed area ( g r e e n ) \color{#20A900}{(green)} between the curves :

  • y = 2 x ( r e d ) y = 2 |x| ~~~~\color{#D61F06}{(red)}
  • y = 8 x 2 ( b l u e ) y = 8 - x^2~~\color{#3D99F6}{(blue)}

If your answer is of the form A B \dfrac{A}{B} , where A A and B B are positive coprime integers , then enter the value of A + B A+B .


The answer is 59.

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1 solution

Fiki Akbar
Nov 14, 2017

Since the graph is symmetric, then we only have to consider the area on the right side of the graph. The green area is the enclosed area between the parabolic curve and x axis minus the area between the line curve and x axis (bottom triangle).

The enclosed area between the parabolic curve and x axis is 0 2 ( 8 x 2 ) d x = 40 3 \int_{0}^{2}\:(8 - x^2)\:dx = \frac{40}{3} and the area between the line curve and x axis (bottom triangle) is 1 2 × 4 × 2 = 4 \frac{1}{2}\times 4\times 2 = 4 Thus, the green area in the right side is 40 3 4 = 28 3 \frac{40}{3} - 4 = \frac{28}{3} .

Then the total area is 2 × 28 3 = 56 3 2\times \frac{28}{3} = \frac{56}{3} . Hence, A + B = 59 A+B = 59 .

Nicely done sir(+1).

Rishu Jaar - 3 years, 6 months ago

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