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Question: Find the volume of the solid obtained by revolving the loop of the curve \(2a{{y}^{2}}=x{{\left( x-a...

Find the volume of the solid obtained by revolving the loop of the curve 2ay2=x(xa)22a{{y}^{2}}=x{{\left( x-a \right)}^{2}} about the x-axis.

Explanation

Solution

Here we have to find the volume of the solid which we will obtain after revolving the loop of the given curve 2ay2=x(xa)22a{{y}^{2}}=x{{\left( x-a \right)}^{2}}. For that, we will plot the graph and we will find the range of xx by putting y=0y=0. Then we will find the volume using integration with proper limits of xx. The value which we will obtain after integration will be the required volume of the solid generated.

Complete step-by-step answer:
The given curve is 2ay2=x(xa)22a{{y}^{2}}=x{{\left( x-a \right)}^{2}}

We will find the value of xx for which y=0y=0. For that, we will put y=0y=0 in the equation of the curve.
Putting value y=0y=0 in the equation, we get
2a02=x(xa)2\Rightarrow 2a{{0}^{2}}=x{{\left( x-a \right)}^{2}}
On simplifying the terms, we get
x=0x=0 and x=ax=a
Now, we will draw the graph of the curve.
Now, we will find the volume of solid obtained by the revolving the loop of this curve.
Therefore,
volume=π0ay2dx\Rightarrow volume=\pi \int\limits_{0}^{a}{{{y}^{2}}dx}
We will put the value of y2{{y}^{2}}here.
volume=π0ax(xa)22adx\Rightarrow volume=\pi \int\limits_{0}^{a}{\dfrac{x{{\left( x-a \right)}^{2}}}{2a}dx}
We will take constants out of integration and we will expand the terms.
volume=π2a0ax(x2+a22ax)dx\Rightarrow volume=\dfrac{\pi }{2a}\int\limits_{0}^{a}{x\left( {{x}^{2}}+{{a}^{2}}-2ax \right)dx}
Multiplying the terms, we get
volume=π2a0a(x3+a2x2ax2)dx\Rightarrow volume=\dfrac{\pi }{2a}\int\limits_{0}^{a}{\left( {{x}^{3}}+{{a}^{2}}{{x}}-2a{{x}^{2}} \right)dx}
Integrating the terms, we get
volume=π2a[x44+a2x222ax33]0a\Rightarrow volume=\dfrac{\pi }{2a}\left[ \dfrac{{{x}^{4}}}{4}+\dfrac{{{a}^{2}}{{x}^{2}}}{2}-\dfrac{2a{{x}^{3}}}{3} \right]_{0}^{a}
On further simplification, we get
volume=π2a[3a4+6a48a412]\Rightarrow volume=\dfrac{\pi }{2a}\left[ \dfrac{3{{a}^{4}}+6{{a}^{4}}-8{{a}^{4}}}{12} \right]
volume=πa324\Rightarrow volume=\dfrac{\pi {{a}^{3}}}{24}
Hence, the required volume of the solid obtained by revolving the loop of this curve is πa324\dfrac{\pi {{a}^{3}}}{24} cubic units.

Note: This curve is symmetric about the x-axis. A curve is said to be symmetric about the x axis if whenever a point (a,b)\left( a,b \right) lies on the curve then point (a,b)\left( a,-b \right) also lies on the curve i.e. both of them will satisfy the equation of the curve. Here x-axis is called the axis of symmetry of the given curve. The shape of the curve is the same on both sides of the axis of symmetry.