Question

Sketch the region lying in the first quadrant and bounded by \[y=9{{x}^{2}}\] , x=0, y=1 and y=4. Find the area of the region using integration.

Answer 1

Hint: Before writing anything, we must know that the y-axis is also known as the x=0 line, so, in the question, the y-axis is being talked about. Use the fact that the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x= b is given by $\int_{a}^{b}{\left| f\left( x \right) \right|dx}$.

Complete step-by-step answer:
Hence, in this question, we are asked to find the area bounded by the curve \[y=4{{x}^{2}}\] and the y-axis which means that the region between the abscissas y=a and y=b, hence we can manipulate the above mentioned fact as follows
The area bounded by the curve x = f(y), the y-axis and the abscissas y = a and y= b is given by $\int_{a}^{b}{\left| f\left( y \right) \right|dy}$.
Now, we can write the curve as function of y as follows
\[\begin{align}
  & \Rightarrow y=\left( 9{{x}^{2}} \right) \\
 & \Rightarrow {{x}^{2}}=\dfrac{y}{49} \\
 & \Rightarrow x=\pm \dfrac{\sqrt{y}}{3} \\
 & \Rightarrow f(y)=\pm \dfrac{\sqrt{y}}{3} \\
\end{align}\]
Now, we can simply take the value of f(y) to be as positive by looking at the figure and also, it is written in the question that we have to consider the first quadrant.
Hence the required area is given by $\int_{a}^{b}{f(y)dy}$.
As mentioned in the question, we have to find the area of the region that is bounded by the curve from the above and the y-axis from the lower side between the two abscissas 1 and 4 which is mentioned in the question itself.

Hence, the values of y are 1 and 4 and these are the values of a and b respectively.
Now, we can simply form the integral as follows to get the required area of the region.
 \[\begin{align}
  & \Rightarrow Area=\int_{a}^{b}{f(y)dy} \\
 & \Rightarrow Area=\int_{1}^{4}{\dfrac{\sqrt{\left( y \right)}}{3}dy} \\
 & \Rightarrow Area=\dfrac{1}{3}\int_{2}^{4}{\sqrt{\left( y \right)}dy} \\
 & \Rightarrow Area=\dfrac{1}{3}\left[ \dfrac{{{\left( y \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]_{1}^{4} \\
 & \Rightarrow Area=\dfrac{1}{3}\left[ \dfrac{{{\left( 4 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{\left( 1 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right] \\
 & \Rightarrow Area=\dfrac{1}{3}\left[ \dfrac{2\left( {{\left( 4 \right)}^{\dfrac{3}{2}}}-1 \right)}{3} \right] \\
 & \Rightarrow Area=\left[ \dfrac{2\left( {{\left( 2 \right)}^{\dfrac{3}{2}\times 2}}-1 \right)}{9} \right] \\
 & \Rightarrow Area=\left[ \dfrac{2\left( {{\left( 2 \right)}^{3}}-1 \right)}{9} \right] \\
 & \Rightarrow Area=\left[ \dfrac{2\left( 8-1 \right)}{9} \right] \\
 & \Rightarrow Area=\left[ \dfrac{14}{9} \right] \\
\end{align}\]
(as we know that the integral of \[{{x}^{n}}\] is as follows
\[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c\]
Where c is the constant of integration )
Hence, we have
Total area \[=\dfrac{14}{9}\] square units

Note: The students can make an error if they don’t know the basic concepts of integration as without knowing them one could never get to the correct answer.
As in this question, it is very important to know the following result beforehand.
The integral of \[{{x}^{n}}\] is as follows
\[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c\]
(Where c is the constant of integration)