Chapter 8: Applications of Integrals

Ans:  Rewrite the given equation of ellipse

\[\frac{{{x}^{2}}}{{{4}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]

The intersection point of the ellipse and the x-axis are at \[x=\pm 4\].

The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].

The given equation of the ellipse can be represented as

If we change \[x\] to \[-x\] or \[y\] to \[y\] the equation remains the same. 

So, the ellipse is symmetrical about the x-axis and the y-axis.

Area bounded by ellipse \[=4\times\] Area of OABO

Area of OABO \[=\int\limits_{0}^{4}{ydx}\]

\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\]

\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{16}\]

\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{16} \right)\]

\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{16}}\]

Area of OABO \[=\int\limits_{0}^{4}{3\sqrt{1-\frac{{{x}^{2}}}{16}}dx}\]

Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}dx}\]

Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}\]

Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation  

Area of OABO \[=\frac{3}{4}\left[ \frac{x}{2}\sqrt{16-{{x}^{2}}}+\frac{16}{2}{{\sin }^{-1}}\frac{x}{4} \right]_{0}^{4}\]

Area of OABO \[=\frac{3}{4}\left[ 0+8{{\sin }^{-1}}(1)-0-8{{\sin }^{-1}}(0) \right]_{0}^{4}\]

Area of OABO \[=\frac{3}{4}\left( \frac{8\pi }{2} \right)\]

Area of OABO \[=3\pi \]

Therefore, the area bounded by ellipse \[=\text{4}\times \text{3}\pi\]

The area bounded by ellipse \[=\text{12}\pi \text{ sq}\text{. Units}\]

Ans: Rewrite the given equation of the ellipse

\[\frac{{{x}^{2}}}{{{2}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]

The intersection point of the ellipse and the x-axis are at \[x=\pm 2\].

The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].

The given equation of the ellipse can be represented as a vertical ellipse.

It can be observed that the ellipse is symmetrical about the x-axis and y-axis.

Since, if we change \[x\] to \[-x\] or \[y\] to \[-y\] the equation remains the same. 

Area bounded by ellipse \[=4\times\]Area of OABO

Area of OABO \[=\int\limits_{0}^{2}{ydx}\]…………………….(A)

\[\Rightarrow \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\]

\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{4}\]

\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{4} \right)\]

\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{4}}\text{ }\]

\[\Rightarrow y=\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }\]…………………..(1)

Area of OABO \[=\int\limits_{0}^{2}{\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }dx}\] using equation (1) in (A)

Area of OABO \[=\frac{3}{2}\int\limits_{0}^{2}{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}\]

Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation  

Area of OABO \[=\frac{3}{2}\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]

Area of OABO \[=\frac{3}{2}\left[ \frac{2\pi }{2} \right]\]

Area of OABO \[=\frac{3\pi }{2}\]

Area bound by the ellipse \[=\text{4}\times \frac{\text{3}\pi }{2}\]

Area bound by the ellipse \[=\text{6}\pi \text{ sq}\text{. Units}\]

Choose the correct answer in the following Exercises 3 and 4.

Ans:

Area of \[OAB=\int_{0}^{2}{ydx}\]

Since, \[{{\text{x}}^2} + {y^2} = 4\]   $ [\text {Equation of circle}]$

\[{{y}^{2}}=4-{{\text{x}}^{2}}\]

\[y=\sqrt{4-{{\text{x}}^{2}}}\]

So,

Area of \[OAB=\int_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\]

Area of \[OAB=\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]

Area of \[OAB=\left[ \frac{2}{2}\sqrt{4-{{2}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{2}{2}-\left( \frac{0}{2}\sqrt{4-{{0}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{0}{2} \right) \right]\]

Area of \[OAB=\left[ \sqrt{4-4}+2{{\sin }^{-1}}1-0 \right]\]

Area of \[OAB=\left[ 2\times \frac{\pi }{2} \right]\]

Area of \[OAB=\pi \] square units

Hence, the area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is \[\pi \] square units.

Hence, option (A) is the correct answer.

Ans:

The area bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is the area of the region \[OAB\].

Area of \[OAB=\int_{0}^{3}{xdy}\]

Area of \[OAB=\int_{0}^{3}{\frac{{{y}^{2}}}{4}dy}\]

Area of \[OAB=\frac{1}{4}\int_{0}^{3}{{{y}^{2}}dy}\]

Area of \[OAB=\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{0}^{3}\]

Area of \[OAB=\frac{1}{4}\left[ \frac{{{3}^{3}}}{3}-0 \right]\]

Area of \[OAB=\frac{1}{4}\left[ \frac{27}{3} \right]\]

Area of \[OAB=\frac{1}{4}\left[ 9 \right]\]

Area of \[OAB=\frac{9}{4}\]square units

Hence, option (B) is the correct answer.