87008
The figure shows a triangle \(A O B\) and the parabola \(y=x^{2}\). The ratio of the area of the triangle \(A O B\) to the area of the region \(A O B\) of the parabola \(y=x^{2}\) is equal to
1 \(\frac{3}{5}\)
2 \(\frac{3}{4}\)
3 \(\frac{7}{8}\)
4 \(\frac{5}{6}\)
5 \(\frac{2}{3}\)
Explanation:
(B) : Given, Area of \(\triangle \mathrm{AOB}=\frac{1}{2} \times\) base \(\times\) height \(=\frac{1}{2} \times 2 \mathrm{a} \times \mathrm{a}^{2}=\mathrm{a}^{3}\) square unit We find the area of region \(\triangle \mathrm{AOB}\) \(=2 \int_{0}^{a^{2}} x d y=2 \int_{0}^{a^{2}} \sqrt{y} d y=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_{0}^{a^{2}}=\frac{4}{3} a^{3}\) square unit \(\therefore\) Ratio of areas \(=\frac{\mathrm{a}^{3}}{\frac{4}{3} \mathrm{a}^{3}}=\frac{3 \mathrm{a}^{3}}{4 \mathrm{a}^{3}}=\frac{3}{4}\)
Kerala CEE-2009
Application of the Integrals
87009
The area of the region described by \(A=\{(x, y)\) : \(\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1\) and \(\left.\mathrm{y}^{2} \leq 1-\mathrm{x}\right\}\) is
1 \(\frac{\pi}{2}+\frac{4}{3}\)
2 \(\frac{\pi}{2}-\frac{4}{3}\)
3 \(\frac{\pi}{2}-\frac{2}{3}\)
4 \(\frac{\pi}{2}+\frac{2}{3}\)
Explanation:
(A) : Given, that the curve, \(x^{2}+y^{2} \leq 1\) and \(y^{2} \leq 1-x\) \(x^{2}+y^{2}=1 \tag{i}\) \(y^{2}=1-x \tag{ii}\) Putting the value of \(y^{2}=1-x\) in equation (i), we get \(x^{2}+1-x=1\) \(x^{2}-x=0\) \(x(x-1)=0\) \(x=0,1\) Putting these value in equation (ii), we get, \(y=0, \pm 1\) So, points of intersection are- \((1,0),(0,1),(0,-1)\) \(\therefore\) Area \(=\) Area of semicircle \(+2 \int_{0}^{1} \sqrt{1-\mathrm{x}} \mathrm{dx}\) \(=\frac{\pi \mathrm{r}^{2}}{2}+2\left[\frac{-(1-\mathrm{x})^{3 / 2}}{3 / 2}\right]_{0}^{1}\) \(=\left(\frac{\pi}{2}+\frac{4}{3}\right)\) equation unit \([\because \mathrm{r}=1]\)
JEE Main-2014
Application of the Integrals
87010
The area (in sq. units) of the largest rectangle \(A B C D\) whose vertices \(A\) and \(B\) lie on the \(X\)-axis and vertices \(C\) and \(D\) lie on the parabola, \(y=x^{2}\) -1 below the \(\mathrm{X}\)-axis, is
87011
The area of the region, enclosed by the circle \(x^{2}\) \(+y^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(y=x\), is
1 \(\frac{1}{3}(12 \pi-1)\)
2 \(\frac{1}{6}(12 \pi-1)\)
3 \(\frac{1}{6}(24 \pi-1)\)
4 \(\frac{1}{3}(6 \pi-1)\)
Explanation:
(B) : Given that the circle, \(x^{2}+y^{2}=2\) Parabola, \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(\mathrm{y}=\mathrm{x}\) \(\therefore\) Required area \(=\) Area of circle - Area bounded by given line and parabola \(=\pi(\sqrt{2})^{2}-\int_{0}^{1}\left(y-y^{2}\right) \mathrm{dy}\) \(=2 \pi-\left[\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{1}=2 \pi-\frac{1}{6}\) \(=\frac{1}{6}(12 \pi-1)\) Sq.unit
JEE Main-2020-08.03.2020
Application of the Integrals
87012
Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{\left|x^{2}\right|}{4}+\frac{y^{2}}{9}=\) 1 is
1 \(6(\pi-2)\)
2 \(3(\pi-2)\)
3 \(3(4-\pi)\)
4 \(6(4-\pi)\)
Explanation:
(A) : Given that, the outside ellipse \(\frac{|x|}{2}+\frac{|y|}{3}=1\) And inside of the ellipse, \(\frac{\left|\mathrm{x}^{2}\right|}{4}+\frac{\left|\mathrm{y}^{2}\right|}{9}=1\) \(\therefore\) Required area \(=\) area of ellipse - Area of quadrilateral \(=\pi \mathrm{ab}-\frac{1}{2} \times 6 \times 4=\pi \times 2 \times 3-12=6(\pi-2)\)
87008
The figure shows a triangle \(A O B\) and the parabola \(y=x^{2}\). The ratio of the area of the triangle \(A O B\) to the area of the region \(A O B\) of the parabola \(y=x^{2}\) is equal to
1 \(\frac{3}{5}\)
2 \(\frac{3}{4}\)
3 \(\frac{7}{8}\)
4 \(\frac{5}{6}\)
5 \(\frac{2}{3}\)
Explanation:
(B) : Given, Area of \(\triangle \mathrm{AOB}=\frac{1}{2} \times\) base \(\times\) height \(=\frac{1}{2} \times 2 \mathrm{a} \times \mathrm{a}^{2}=\mathrm{a}^{3}\) square unit We find the area of region \(\triangle \mathrm{AOB}\) \(=2 \int_{0}^{a^{2}} x d y=2 \int_{0}^{a^{2}} \sqrt{y} d y=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_{0}^{a^{2}}=\frac{4}{3} a^{3}\) square unit \(\therefore\) Ratio of areas \(=\frac{\mathrm{a}^{3}}{\frac{4}{3} \mathrm{a}^{3}}=\frac{3 \mathrm{a}^{3}}{4 \mathrm{a}^{3}}=\frac{3}{4}\)
Kerala CEE-2009
Application of the Integrals
87009
The area of the region described by \(A=\{(x, y)\) : \(\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1\) and \(\left.\mathrm{y}^{2} \leq 1-\mathrm{x}\right\}\) is
1 \(\frac{\pi}{2}+\frac{4}{3}\)
2 \(\frac{\pi}{2}-\frac{4}{3}\)
3 \(\frac{\pi}{2}-\frac{2}{3}\)
4 \(\frac{\pi}{2}+\frac{2}{3}\)
Explanation:
(A) : Given, that the curve, \(x^{2}+y^{2} \leq 1\) and \(y^{2} \leq 1-x\) \(x^{2}+y^{2}=1 \tag{i}\) \(y^{2}=1-x \tag{ii}\) Putting the value of \(y^{2}=1-x\) in equation (i), we get \(x^{2}+1-x=1\) \(x^{2}-x=0\) \(x(x-1)=0\) \(x=0,1\) Putting these value in equation (ii), we get, \(y=0, \pm 1\) So, points of intersection are- \((1,0),(0,1),(0,-1)\) \(\therefore\) Area \(=\) Area of semicircle \(+2 \int_{0}^{1} \sqrt{1-\mathrm{x}} \mathrm{dx}\) \(=\frac{\pi \mathrm{r}^{2}}{2}+2\left[\frac{-(1-\mathrm{x})^{3 / 2}}{3 / 2}\right]_{0}^{1}\) \(=\left(\frac{\pi}{2}+\frac{4}{3}\right)\) equation unit \([\because \mathrm{r}=1]\)
JEE Main-2014
Application of the Integrals
87010
The area (in sq. units) of the largest rectangle \(A B C D\) whose vertices \(A\) and \(B\) lie on the \(X\)-axis and vertices \(C\) and \(D\) lie on the parabola, \(y=x^{2}\) -1 below the \(\mathrm{X}\)-axis, is
87011
The area of the region, enclosed by the circle \(x^{2}\) \(+y^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(y=x\), is
1 \(\frac{1}{3}(12 \pi-1)\)
2 \(\frac{1}{6}(12 \pi-1)\)
3 \(\frac{1}{6}(24 \pi-1)\)
4 \(\frac{1}{3}(6 \pi-1)\)
Explanation:
(B) : Given that the circle, \(x^{2}+y^{2}=2\) Parabola, \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(\mathrm{y}=\mathrm{x}\) \(\therefore\) Required area \(=\) Area of circle - Area bounded by given line and parabola \(=\pi(\sqrt{2})^{2}-\int_{0}^{1}\left(y-y^{2}\right) \mathrm{dy}\) \(=2 \pi-\left[\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{1}=2 \pi-\frac{1}{6}\) \(=\frac{1}{6}(12 \pi-1)\) Sq.unit
JEE Main-2020-08.03.2020
Application of the Integrals
87012
Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{\left|x^{2}\right|}{4}+\frac{y^{2}}{9}=\) 1 is
1 \(6(\pi-2)\)
2 \(3(\pi-2)\)
3 \(3(4-\pi)\)
4 \(6(4-\pi)\)
Explanation:
(A) : Given that, the outside ellipse \(\frac{|x|}{2}+\frac{|y|}{3}=1\) And inside of the ellipse, \(\frac{\left|\mathrm{x}^{2}\right|}{4}+\frac{\left|\mathrm{y}^{2}\right|}{9}=1\) \(\therefore\) Required area \(=\) area of ellipse - Area of quadrilateral \(=\pi \mathrm{ab}-\frac{1}{2} \times 6 \times 4=\pi \times 2 \times 3-12=6(\pi-2)\)
87008
The figure shows a triangle \(A O B\) and the parabola \(y=x^{2}\). The ratio of the area of the triangle \(A O B\) to the area of the region \(A O B\) of the parabola \(y=x^{2}\) is equal to
1 \(\frac{3}{5}\)
2 \(\frac{3}{4}\)
3 \(\frac{7}{8}\)
4 \(\frac{5}{6}\)
5 \(\frac{2}{3}\)
Explanation:
(B) : Given, Area of \(\triangle \mathrm{AOB}=\frac{1}{2} \times\) base \(\times\) height \(=\frac{1}{2} \times 2 \mathrm{a} \times \mathrm{a}^{2}=\mathrm{a}^{3}\) square unit We find the area of region \(\triangle \mathrm{AOB}\) \(=2 \int_{0}^{a^{2}} x d y=2 \int_{0}^{a^{2}} \sqrt{y} d y=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_{0}^{a^{2}}=\frac{4}{3} a^{3}\) square unit \(\therefore\) Ratio of areas \(=\frac{\mathrm{a}^{3}}{\frac{4}{3} \mathrm{a}^{3}}=\frac{3 \mathrm{a}^{3}}{4 \mathrm{a}^{3}}=\frac{3}{4}\)
Kerala CEE-2009
Application of the Integrals
87009
The area of the region described by \(A=\{(x, y)\) : \(\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1\) and \(\left.\mathrm{y}^{2} \leq 1-\mathrm{x}\right\}\) is
1 \(\frac{\pi}{2}+\frac{4}{3}\)
2 \(\frac{\pi}{2}-\frac{4}{3}\)
3 \(\frac{\pi}{2}-\frac{2}{3}\)
4 \(\frac{\pi}{2}+\frac{2}{3}\)
Explanation:
(A) : Given, that the curve, \(x^{2}+y^{2} \leq 1\) and \(y^{2} \leq 1-x\) \(x^{2}+y^{2}=1 \tag{i}\) \(y^{2}=1-x \tag{ii}\) Putting the value of \(y^{2}=1-x\) in equation (i), we get \(x^{2}+1-x=1\) \(x^{2}-x=0\) \(x(x-1)=0\) \(x=0,1\) Putting these value in equation (ii), we get, \(y=0, \pm 1\) So, points of intersection are- \((1,0),(0,1),(0,-1)\) \(\therefore\) Area \(=\) Area of semicircle \(+2 \int_{0}^{1} \sqrt{1-\mathrm{x}} \mathrm{dx}\) \(=\frac{\pi \mathrm{r}^{2}}{2}+2\left[\frac{-(1-\mathrm{x})^{3 / 2}}{3 / 2}\right]_{0}^{1}\) \(=\left(\frac{\pi}{2}+\frac{4}{3}\right)\) equation unit \([\because \mathrm{r}=1]\)
JEE Main-2014
Application of the Integrals
87010
The area (in sq. units) of the largest rectangle \(A B C D\) whose vertices \(A\) and \(B\) lie on the \(X\)-axis and vertices \(C\) and \(D\) lie on the parabola, \(y=x^{2}\) -1 below the \(\mathrm{X}\)-axis, is
87011
The area of the region, enclosed by the circle \(x^{2}\) \(+y^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(y=x\), is
1 \(\frac{1}{3}(12 \pi-1)\)
2 \(\frac{1}{6}(12 \pi-1)\)
3 \(\frac{1}{6}(24 \pi-1)\)
4 \(\frac{1}{3}(6 \pi-1)\)
Explanation:
(B) : Given that the circle, \(x^{2}+y^{2}=2\) Parabola, \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(\mathrm{y}=\mathrm{x}\) \(\therefore\) Required area \(=\) Area of circle - Area bounded by given line and parabola \(=\pi(\sqrt{2})^{2}-\int_{0}^{1}\left(y-y^{2}\right) \mathrm{dy}\) \(=2 \pi-\left[\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{1}=2 \pi-\frac{1}{6}\) \(=\frac{1}{6}(12 \pi-1)\) Sq.unit
JEE Main-2020-08.03.2020
Application of the Integrals
87012
Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{\left|x^{2}\right|}{4}+\frac{y^{2}}{9}=\) 1 is
1 \(6(\pi-2)\)
2 \(3(\pi-2)\)
3 \(3(4-\pi)\)
4 \(6(4-\pi)\)
Explanation:
(A) : Given that, the outside ellipse \(\frac{|x|}{2}+\frac{|y|}{3}=1\) And inside of the ellipse, \(\frac{\left|\mathrm{x}^{2}\right|}{4}+\frac{\left|\mathrm{y}^{2}\right|}{9}=1\) \(\therefore\) Required area \(=\) area of ellipse - Area of quadrilateral \(=\pi \mathrm{ab}-\frac{1}{2} \times 6 \times 4=\pi \times 2 \times 3-12=6(\pi-2)\)
87008
The figure shows a triangle \(A O B\) and the parabola \(y=x^{2}\). The ratio of the area of the triangle \(A O B\) to the area of the region \(A O B\) of the parabola \(y=x^{2}\) is equal to
1 \(\frac{3}{5}\)
2 \(\frac{3}{4}\)
3 \(\frac{7}{8}\)
4 \(\frac{5}{6}\)
5 \(\frac{2}{3}\)
Explanation:
(B) : Given, Area of \(\triangle \mathrm{AOB}=\frac{1}{2} \times\) base \(\times\) height \(=\frac{1}{2} \times 2 \mathrm{a} \times \mathrm{a}^{2}=\mathrm{a}^{3}\) square unit We find the area of region \(\triangle \mathrm{AOB}\) \(=2 \int_{0}^{a^{2}} x d y=2 \int_{0}^{a^{2}} \sqrt{y} d y=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_{0}^{a^{2}}=\frac{4}{3} a^{3}\) square unit \(\therefore\) Ratio of areas \(=\frac{\mathrm{a}^{3}}{\frac{4}{3} \mathrm{a}^{3}}=\frac{3 \mathrm{a}^{3}}{4 \mathrm{a}^{3}}=\frac{3}{4}\)
Kerala CEE-2009
Application of the Integrals
87009
The area of the region described by \(A=\{(x, y)\) : \(\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1\) and \(\left.\mathrm{y}^{2} \leq 1-\mathrm{x}\right\}\) is
1 \(\frac{\pi}{2}+\frac{4}{3}\)
2 \(\frac{\pi}{2}-\frac{4}{3}\)
3 \(\frac{\pi}{2}-\frac{2}{3}\)
4 \(\frac{\pi}{2}+\frac{2}{3}\)
Explanation:
(A) : Given, that the curve, \(x^{2}+y^{2} \leq 1\) and \(y^{2} \leq 1-x\) \(x^{2}+y^{2}=1 \tag{i}\) \(y^{2}=1-x \tag{ii}\) Putting the value of \(y^{2}=1-x\) in equation (i), we get \(x^{2}+1-x=1\) \(x^{2}-x=0\) \(x(x-1)=0\) \(x=0,1\) Putting these value in equation (ii), we get, \(y=0, \pm 1\) So, points of intersection are- \((1,0),(0,1),(0,-1)\) \(\therefore\) Area \(=\) Area of semicircle \(+2 \int_{0}^{1} \sqrt{1-\mathrm{x}} \mathrm{dx}\) \(=\frac{\pi \mathrm{r}^{2}}{2}+2\left[\frac{-(1-\mathrm{x})^{3 / 2}}{3 / 2}\right]_{0}^{1}\) \(=\left(\frac{\pi}{2}+\frac{4}{3}\right)\) equation unit \([\because \mathrm{r}=1]\)
JEE Main-2014
Application of the Integrals
87010
The area (in sq. units) of the largest rectangle \(A B C D\) whose vertices \(A\) and \(B\) lie on the \(X\)-axis and vertices \(C\) and \(D\) lie on the parabola, \(y=x^{2}\) -1 below the \(\mathrm{X}\)-axis, is
87011
The area of the region, enclosed by the circle \(x^{2}\) \(+y^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(y=x\), is
1 \(\frac{1}{3}(12 \pi-1)\)
2 \(\frac{1}{6}(12 \pi-1)\)
3 \(\frac{1}{6}(24 \pi-1)\)
4 \(\frac{1}{3}(6 \pi-1)\)
Explanation:
(B) : Given that the circle, \(x^{2}+y^{2}=2\) Parabola, \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(\mathrm{y}=\mathrm{x}\) \(\therefore\) Required area \(=\) Area of circle - Area bounded by given line and parabola \(=\pi(\sqrt{2})^{2}-\int_{0}^{1}\left(y-y^{2}\right) \mathrm{dy}\) \(=2 \pi-\left[\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{1}=2 \pi-\frac{1}{6}\) \(=\frac{1}{6}(12 \pi-1)\) Sq.unit
JEE Main-2020-08.03.2020
Application of the Integrals
87012
Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{\left|x^{2}\right|}{4}+\frac{y^{2}}{9}=\) 1 is
1 \(6(\pi-2)\)
2 \(3(\pi-2)\)
3 \(3(4-\pi)\)
4 \(6(4-\pi)\)
Explanation:
(A) : Given that, the outside ellipse \(\frac{|x|}{2}+\frac{|y|}{3}=1\) And inside of the ellipse, \(\frac{\left|\mathrm{x}^{2}\right|}{4}+\frac{\left|\mathrm{y}^{2}\right|}{9}=1\) \(\therefore\) Required area \(=\) area of ellipse - Area of quadrilateral \(=\pi \mathrm{ab}-\frac{1}{2} \times 6 \times 4=\pi \times 2 \times 3-12=6(\pi-2)\)
87008
The figure shows a triangle \(A O B\) and the parabola \(y=x^{2}\). The ratio of the area of the triangle \(A O B\) to the area of the region \(A O B\) of the parabola \(y=x^{2}\) is equal to
1 \(\frac{3}{5}\)
2 \(\frac{3}{4}\)
3 \(\frac{7}{8}\)
4 \(\frac{5}{6}\)
5 \(\frac{2}{3}\)
Explanation:
(B) : Given, Area of \(\triangle \mathrm{AOB}=\frac{1}{2} \times\) base \(\times\) height \(=\frac{1}{2} \times 2 \mathrm{a} \times \mathrm{a}^{2}=\mathrm{a}^{3}\) square unit We find the area of region \(\triangle \mathrm{AOB}\) \(=2 \int_{0}^{a^{2}} x d y=2 \int_{0}^{a^{2}} \sqrt{y} d y=2\left[\frac{y^{3 / 2}}{3 / 2}\right]_{0}^{a^{2}}=\frac{4}{3} a^{3}\) square unit \(\therefore\) Ratio of areas \(=\frac{\mathrm{a}^{3}}{\frac{4}{3} \mathrm{a}^{3}}=\frac{3 \mathrm{a}^{3}}{4 \mathrm{a}^{3}}=\frac{3}{4}\)
Kerala CEE-2009
Application of the Integrals
87009
The area of the region described by \(A=\{(x, y)\) : \(\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1\) and \(\left.\mathrm{y}^{2} \leq 1-\mathrm{x}\right\}\) is
1 \(\frac{\pi}{2}+\frac{4}{3}\)
2 \(\frac{\pi}{2}-\frac{4}{3}\)
3 \(\frac{\pi}{2}-\frac{2}{3}\)
4 \(\frac{\pi}{2}+\frac{2}{3}\)
Explanation:
(A) : Given, that the curve, \(x^{2}+y^{2} \leq 1\) and \(y^{2} \leq 1-x\) \(x^{2}+y^{2}=1 \tag{i}\) \(y^{2}=1-x \tag{ii}\) Putting the value of \(y^{2}=1-x\) in equation (i), we get \(x^{2}+1-x=1\) \(x^{2}-x=0\) \(x(x-1)=0\) \(x=0,1\) Putting these value in equation (ii), we get, \(y=0, \pm 1\) So, points of intersection are- \((1,0),(0,1),(0,-1)\) \(\therefore\) Area \(=\) Area of semicircle \(+2 \int_{0}^{1} \sqrt{1-\mathrm{x}} \mathrm{dx}\) \(=\frac{\pi \mathrm{r}^{2}}{2}+2\left[\frac{-(1-\mathrm{x})^{3 / 2}}{3 / 2}\right]_{0}^{1}\) \(=\left(\frac{\pi}{2}+\frac{4}{3}\right)\) equation unit \([\because \mathrm{r}=1]\)
JEE Main-2014
Application of the Integrals
87010
The area (in sq. units) of the largest rectangle \(A B C D\) whose vertices \(A\) and \(B\) lie on the \(X\)-axis and vertices \(C\) and \(D\) lie on the parabola, \(y=x^{2}\) -1 below the \(\mathrm{X}\)-axis, is
87011
The area of the region, enclosed by the circle \(x^{2}\) \(+y^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(y=x\), is
1 \(\frac{1}{3}(12 \pi-1)\)
2 \(\frac{1}{6}(12 \pi-1)\)
3 \(\frac{1}{6}(24 \pi-1)\)
4 \(\frac{1}{3}(6 \pi-1)\)
Explanation:
(B) : Given that the circle, \(x^{2}+y^{2}=2\) Parabola, \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(\mathrm{y}=\mathrm{x}\) \(\therefore\) Required area \(=\) Area of circle - Area bounded by given line and parabola \(=\pi(\sqrt{2})^{2}-\int_{0}^{1}\left(y-y^{2}\right) \mathrm{dy}\) \(=2 \pi-\left[\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{1}=2 \pi-\frac{1}{6}\) \(=\frac{1}{6}(12 \pi-1)\) Sq.unit
JEE Main-2020-08.03.2020
Application of the Integrals
87012
Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{\left|x^{2}\right|}{4}+\frac{y^{2}}{9}=\) 1 is
1 \(6(\pi-2)\)
2 \(3(\pi-2)\)
3 \(3(4-\pi)\)
4 \(6(4-\pi)\)
Explanation:
(A) : Given that, the outside ellipse \(\frac{|x|}{2}+\frac{|y|}{3}=1\) And inside of the ellipse, \(\frac{\left|\mathrm{x}^{2}\right|}{4}+\frac{\left|\mathrm{y}^{2}\right|}{9}=1\) \(\therefore\) Required area \(=\) area of ellipse - Area of quadrilateral \(=\pi \mathrm{ab}-\frac{1}{2} \times 6 \times 4=\pi \times 2 \times 3-12=6(\pi-2)\)
