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Question: If the area enclosed between the curves \(y=k{{x}^{2}}\) and \(x=k{{y}^{2}}\), (k > 0), is 1 square ...

If the area enclosed between the curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}}, (k > 0), is 1 square unit. Then k is?

Explanation

Solution

We start solving this problem by considering the given two curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}}. Then we solve both equations to get the intersecting point of the curves. Then we integrate them to get the area enclosed between them by using the formula ∫[f(y)−f(x)]dx\int{\left[ f\left( y \right)-f\left( x \right) \right]dx} where f(y)f\left( y \right) is the upper curve and f(x)f\left( x \right) is the lower curve. Then we equate this area to 1 and hence we get the value of k.

Complete step-by-step solution:

Let us consider the given curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}}, we get,

x=ky2x=k{{y}^{2}}

⇒x=k(kx2)2\Rightarrow x=k{{\left( k{{x}^{2}} \right)}^{2}}

⇒x=k(k2x4)\Rightarrow x=k\left( {{k}^{2}}{{x}^{4}} \right)

⇒x=k3x4\Rightarrow x={{k}^{3}}{{x}^{4}}

⇒1x3=k3\Rightarrow \dfrac{1}{{{x}^{3}}}={{k}^{3}}

⇒x3=1k3\Rightarrow {{x}^{3}}=\dfrac{1}{{{k}^{3}}}

⇒x=1k\Rightarrow x=\dfrac{1}{k}

Now, we substitute the value of xx in the equation y=kx2y=k{{x}^{2}} , we get,

y=kx2y=k{{x}^{2}}

⇒y=k(1k)2\Rightarrow y=k{{\left( \dfrac{1}{k} \right)}^{2}}

⇒y=kk2\Rightarrow y=\dfrac{k}{{{k}^{2}}}

⇒y=1k\Rightarrow y=\dfrac{1}{k}

So, the point of intersection of the given two curves is (1k,1k)\left( \dfrac{1}{k},\dfrac{1}{k} \right).


Now, we have to find the area of the shaded region.

Let us consider the formula ∫[f(y)−f(x)]dx\int{\left[ f\left( y \right)-f\left( x \right) \right]dx} , limits between the points of intersection of the curves f(y)f\left( y \right) and f(x)f\left( x \right) where f(y)f\left( y \right) is the upper curve and f(x)f\left( x \right) is the lower curve.

By using the above formula, we get,

Area enclosed between the curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}} is given by

A=∫01k(xk−kx2)dxA=\int\limits_{0}^{\frac{1}{k}}{\left( \sqrt{\dfrac{x}{k}}-k{{x}^{2}} \right)dx}

⇒A=[1k.23.x32−k.x33]01k\Rightarrow A=\left[ \dfrac{1}{\sqrt{k}}.\dfrac{2}{3}.{{x}^{\frac{3}{2}}}-k.\dfrac{{{x}^{3}}}{3} \right]_{0}^{\frac{1}{k}}

⇒A=(1k.23.(1k)32−k3.(1k)3)−(1k.23.(0)32−k3.(0)3)\Rightarrow A=\left( \dfrac{1}{\sqrt{k}}.\dfrac{2}{3}.{{\left( \dfrac{1}{k} \right)}^{\frac{3}{2}}}-\dfrac{k}{3}.{{\left( \dfrac{1}{k} \right)}^{3}} \right)-\left( \dfrac{1}{\sqrt{k}}.\dfrac{2}{3}.{{\left( 0 \right)}^{\frac{3}{2}}}-\dfrac{k}{3}.{{\left( 0 \right)}^{3}} \right)

⇒A=23k2−13k2−0\Rightarrow A=\dfrac{2}{3{{k}^{2}}}-\dfrac{1}{3{{k}^{2}}}-0

⇒A=13k2\Rightarrow A=\dfrac{1}{3{{k}^{2}}}

So, Area enclosed between the curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}} is 13k2\dfrac{1}{3{{k}^{2}}}

But we were given that the Area enclosed between the curves y=kx2y=k{{x}^{2}} and x=ky2x=k{{y}^{2}} is 1 square unit.

So, we get,

13k2=1\dfrac{1}{3{{k}^{2}}}=1

⇒3k2=1\Rightarrow 3{{k}^{2}}=1

⇒k2=13\Rightarrow {{k}^{2}}=\dfrac{1}{3}

⇒k=13\Rightarrow k=\sqrt{\dfrac{1}{3}}

Therefore, the value of kk is 13\sqrt{\dfrac{1}{3}}.

Note: The possibilities for making mistakes in this type of problem are one may make a mistake by taking the limits of the integral in reverse order. One can also make a mistake by considering the lower curve as the upper curve and the upper curve as the lower curve while applying the formula of the area enclosed between two curves. Then we will get the answer as negative as our answer.