Question

How do you find the area enclosed by y=sinx and the x-axis for \[0\le x\le \pi \] and the volume of the solid of revolution, when this area is rotated about the x-axis?

Answer 1

Hint: This type of question is based on the concept of application of integration. We are given the equation y=sinx in the interval \[0\le x\le \pi \]. To find the area enclosed by a curve in a given interval, we have to integrate the equation with the intervals as the limits of integration. Here, we have to integrate y with respect to x with the limits 0 and \[\pi \]. We know that \[\int{\sin xdx=-\cos x}\]. Using this, we can find the definite integral of the equation which is the area enclosed by the curve. To find the volume about the x-axis, we have to use the formula \[V=\pi \int{{{y}^{2}}}dx\] where V is the volume and the limits are 0 and \[\pi \]. Use trigonometric identity, that is \[{{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}\] and find the integration which is the required answer.

Complete step by step solution:
According to the question, we are asked to find the area enclosed by y=sinx and the x-axis for \[0\le x\le \pi \] and the volume of the solid of revolution, when this area is rotated about the x-axis.
We have been given the equation of the line is y=sinx in the interval \[\left[ 0,\pi \right]\]. -----------(1)
Let us first find the area enclosed by curve (1).
We know that the area enclosed by a curve is the integral of the curve within the limits.
Here, we have been given the function and limit is with respect to x.
Therefore, we have to integrate with respect to x within the limits \[\left[ 0,\pi \right]\].
Therefore, the area A is
\[A=\int\limits_{0}^{\pi }{ydx}\]
\[\Rightarrow A=\int\limits_{0}^{\pi }{\sin xdx}\]
But we know that \[\int{\sin xdx=-\cos x}\]. On substituting to find the area, we get
\[A=\left[ -\cos x \right]_{0}^{\pi }\]
On substituting the limits in place of x, we get
\[A=-\cos \pi -\left( -\cos 0 \right)\]
On further simplification, we get
\[A=-\cos \pi +\cos 0\]
We know that the value of cos0 is 1 and the value of \[\cos \pi =-1\]. On substituting the values in the area, we get
\[A=-\left( -1 \right)+1\]
\[\Rightarrow A=1+1\]
On further simplification, we get
\[A=2\]
Therefore, the area of the curve y=sinx enclosed in the region \[0\le x\le \pi \] is 2 square units.
Now, we have to find the volume of the solid of revolution, when this area is rotated about the x-axis.
We know that the formula to find the volume of a curve enclosed in a given interval is
\[V=\pi \int{{{y}^{2}}}dx\]
We know that y=sinx.
Therefore, we get \[{{y}^{2}}={{\sin }^{2}}x\].
The volume of the considered solid of revolution in the interval \[\left[ 0,\pi \right]\] is
\[V=\pi \int\limits_{0}^{\pi }{{{\sin }^{2}}x}dx\]
We know that \[2{{\sin }^{2}}\theta =1-\cos 2\theta \]. But we need the value of \[{{\sin }^{2}}\theta \].
Therefore, we get \[{{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}\].
Using this identity in the volume, we get
\[V=\pi \int\limits_{0}^{\pi }{\left( \dfrac{1-\cos 2x}{2} \right)}dx\]
\[\Rightarrow V=\dfrac{1}{2}\pi \int\limits_{0}^{\pi }{\left( 1-\cos 2x \right)}dx\]
Using the subtraction rule of integration, that is \[\int{\left( u-v \right)}dx=\int{udx-\int{vdx}}\], we get
\[V=\dfrac{1}{2}\pi \left[ \int\limits_{0}^{\pi }{dx-\int\limits_{0}^{\pi }{\cos 2x}}dx \right]\]
We know that the integral of cos2x is \[\dfrac{\sin 2x}{2}\].
And also using the power rule of integration that is \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\], we get
\[\int{dx}=\dfrac{{{x}^{0+1}}}{0+1}\]
On further simplification, we get
\[\int{dx}=x\]
Therefore, the volume of the solid is
\[V=\dfrac{1}{2}\pi \left\{ \left[ x \right]_{0}^{\pi }-\left[ \dfrac{\cos 2x}{2} \right]_{0}^{\pi } \right\}\]
On substituting the limits to the variable x, we get
\[V=\dfrac{1}{2}\pi \left\{ \left[ \pi -0 \right]-\left[ \dfrac{\cos 2\pi }{2}-\dfrac{\cos 0}{2} \right] \right\}\]
\[\Rightarrow V=\dfrac{1}{2}\pi \left\{ \pi -\left[ \dfrac{\cos 2\pi }{2}-\dfrac{\cos 0}{2} \right] \right\}\]
We know that \[\cos 0=1\] and \[\cos 2\pi =1\]. On substituting the values, we get
\[V=\dfrac{1}{2}\pi \left\{ \pi -\left[ \dfrac{1}{2}-\dfrac{1}{2} \right] \right\}\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling \[\dfrac{1}{2}\], we get
\[V=\dfrac{1}{2}\pi \left( \pi -0 \right)\]
\[\Rightarrow V=\dfrac{1}{2}\pi \left( \pi \right)\]
On further simplifications, we get
\[V=\dfrac{1}{2}{{\pi }^{2}}\]
Therefore, the volume of the solid is \[\dfrac{1}{2}{{\pi }^{2}}\] cubic units.
Hence, the area enclosed by the curve y=sinx in the limits \[0\le x\le \pi \] is 2 sq. units and the volume enclosed by the solid of revolution when the area is rotated about the x-axis is \[\dfrac{1}{2}{{\pi }^{2}}\] cubic units.

Note: Whenever we get such types of problems, we have to use definite integrals. Avoid calculation mistakes based on sign convention. We should know the integration of trigonometric functions to solve this question. We should not forget to put units after finding the area and volume without which the answer is incomplete.