Question

Find the area of the region in the first quadrant enclosed by the x-axis, the line \[y=x\]and the circle\[{{x}^{2}}+{{y}^{2}}=32\].

Answer 1

Hint: First try to make a rough sketch. Then find out the symmetry. Then find the area under the curve using integral.

Complete step-by-step answer:
Consider the circle,

 \[{{x}^{2}}+{{y}^{2}}=32\]

This can be written as,

\[\begin{align}

  & {{x}^{2}}+{{y}^{2}}=\sqrt{16\times 2} \\

 & {{x}^{2}}+{{y}^{2}}=4\sqrt{2} \\

\end{align}\]

Here \[4\sqrt{2}\]is the radius of the given circle.

It is given that we need to find the area enclosed by circle, the x-axis and the line ‘y=x’
First, plot the graph of the region.

The shaded region is the required area.

To find the point of intersection of line ‘y=x’ and\[{{x}^{2}}+{{y}^{2}}=32\]. So we will substitute ‘y=x’ in the equation of circle, we get

\[{{y}^{2}}+{{y}^{2}}=32\]

\[\begin{align}

  & \Rightarrow 2{{y}^{2}}=32 \\

 & \Rightarrow {{y}^{2}}=\dfrac{32}{2}=16 \\

\end{align}\]

Taking square root on both sides, we get

\[\Rightarrow y=\pm 4\]

But we know in the line ‘y=x’, so

\[\Rightarrow x=\pm 4\]

Hence the point of intersection of the given line and circle is (-4, -4) and (4, 4).

As the point is in the first quadrant, the point is (4, 4).

From the figure we can get the required area by finding out the area under the line from ‘x=0’ and ‘x=4’, then add the area under the circle curve from ‘x=4’ to $'x=\sqrt{32}'$.

The formula of finding the area enclosed by $f(x)$ between $x=a$ and $x=b$ can be written as $=|\int\limits_{a}^{b}{f(x)dx|}$.

So the required area under the curve is,

\[\text{Area =}\int\limits_{0}^{4}{x\text{ }dx}+\int\limits_{4}^{\sqrt{32}}{\sqrt{32-{{x}^{2}}}dx}........(i)\]

Now we integrate, \[\sqrt{32-{{x}^{2}}}\], separately.

Substitute,

$\begin{align}

  & x=\sqrt{32}\sin (u) \\

 & \Rightarrow u={{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \\

\end{align}$

Differentiating on both sides, we get

$dx=\sqrt{32}\cos (u)du$

So, we can write,

$\int{\sqrt{32-{{x}^{2}}}}dx=\int{\sqrt{32-{{\left( \sqrt{32}\sin (u) \right)}^{2}}}\left( \sqrt{32}\cos (u)du \right)}$

On solving, we get

$\begin{align}

  & \int{\sqrt{32-{{x}^{2}}}}dx=\int{\sqrt{32-\left( 32{{\sin }^{2}}(u) \right)}\left( \sqrt{32}\cos (u)du \right)} \\

 & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=\int{\sqrt{32\left( 1-{{\sin }^{2}}(u) \right)}\left( \sqrt{32}\cos (u)du \right)} \\

\end{align}$

But we know, $\sqrt{1-{{\sin }^{2}}x}=\cos x$, so the above equation becomes,

$\begin{align}

  & \int{\sqrt{32-{{x}^{2}}}}dx=\int{\sqrt{32}\left( \cos (u) \right)\left( \sqrt{32}\cos (u)du \right)} \\

 & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=\int{32\left( {{\cos }^{2}}(u) \right)du} \\

\end{align}$

Applying reduction formula, we have $\int{{{\cos }^{n}}x}dx=\dfrac{n-1}{n}\int{{{\cos }^{(n-2)}}xdx+\dfrac{{{\cos }^{(n-1)}}x\sin x}{n}}$ , so we get

$\begin{align}

  & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=32\left[ \dfrac{2-1}{2}\int{{{\cos }^{(2-2)}}(u)dx+\dfrac{{{\cos }^{(2-1)}}(u)\sin (u)}{2}} \right] \\

 & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=32\left[ \dfrac{1}{2}\int{1du+\dfrac{\cos (u)\sin (u)}{2}} \right] \\

\end{align}$

Taking out the common term, we get

$\begin{align}

  & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=\dfrac{32}{2}\left[ \int{1du+\cos (u)\sin (u)} \right] \\

 & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=16\left[ u+\cos (u)\sin (u) \right] \\

\end{align}$

Substituting back the value of ‘u’, we get

\[\begin{align}

  & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=16\left[ {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right)+\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \right)\sin \left( {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \right) \right] \\

 & \Rightarrow \int{\sqrt{32-{{x}^{2}}}}dx=16\left[ {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right)+\left( \dfrac{x}{\sqrt{32}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \right) \right] \\

\end{align}\]

Substituting this value in equation (i), we get

\[\text{Area =}\int\limits_{0}^{4}{x\text{ }dx}+16\left[ {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right)+\left( \dfrac{x}{\sqrt{32}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \right) \right]_{4}^{\sqrt{32}}\]

On integrating the first part, we get

\[\text{Area =}\left| \left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{4}+16\left[ {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right)+\left( \dfrac{x}{\sqrt{32}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{x}{\sqrt{32}} \right) \right) \right] \right|_{4}^{\sqrt{32}}\]

Applying the limits, we get

\[\text{Area =}\left| \left[ \dfrac{{{4}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]+16\left[ {{\sin }^{-1}}\left( \dfrac{\sqrt{32}}{\sqrt{32}} \right)+\left( \dfrac{\sqrt{32}}{\sqrt{32}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{32}}{\sqrt{32}} \right) \right) \right]-16\left[ {{\sin }^{-1}}\left( \dfrac{4}{\sqrt{32}} \right)+\left( \dfrac{4}{\sqrt{32}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{4}{\sqrt{32}} \right) \right) \right] \right|\]

Solving this, we get

\[\text{Area =}\left| \left[ 8-0 \right]+16\left[ {{\sin }^{-1}}\left( 1 \right)+\left( 1 \right)\cos \left( {{\sin }^{-1}}\left( 1 \right) \right) \right]-16\left[ {{\sin }^{-1}}\left( \dfrac{4}{4\sqrt{2}} \right)+\left( \dfrac{4}{4\sqrt{2}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{4}{4\sqrt{2}} \right) \right) \right] \right|\]

But, we know ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$ , so above equation becomes,

\[\text{Area =}\left| 8+16\left[ \dfrac{\pi }{2}+\left( 1 \right)\cos \left( \dfrac{\pi }{2} \right) \right]-16\left[ {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{1}{\sqrt{2}} \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \right) \right] \right|\]

But, we know \[{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{4},\cos \left( \dfrac{\pi }{2} \right)=0\] , so above equation becomes,

\[\text{Area =}\left| 8+16\left[ \dfrac{\pi }{2}+0 \right]-16\left[ \dfrac{\pi }{4}+\left( \dfrac{1}{\sqrt{2}} \right)\cos \left( \dfrac{\pi }{4} \right) \right] \right|\]

But, we know \[\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\] , so above equation becomes,

\[\begin{align}

  & \text{Area =}\left| 8+8\pi -16\left[ \dfrac{\pi }{4}+\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right) \right] \right| \\

 & \Rightarrow \text{Area =}\left| 8+8\pi -16\left[ \dfrac{\pi }{4}+\dfrac{1}{2} \right] \right| \\

 & \Rightarrow \text{Area =}\left| 8+8\pi -16\left[ \dfrac{\pi +2}{4} \right] \right| \\

 & \Rightarrow \text{Area =}\left| 8+8\pi -4\pi -8 \right| \\

 & \Rightarrow \text{Area=4}\pi \\

\end{align}\]

So, the area of the region in the first quadrant enclosed by the x-axis, the line \[y=x\]and the circle\[{{x}^{2}}+{{y}^{2}}=32\] is $4\pi $ square. units.


Note: The possibility of mistake is that students might try to calculate the area between the circle, line and y-axis instead.

Students can go wrong when substituting the upper and lower bounds in the integral and calculating the answer.