86999
The area bounded by the parabolas \(y=4 x^{2}\), \(y=\frac{x^{2}}{9}\) and the line \(y=2\) is
1 \(\frac{5 \sqrt{2}}{3}\) sq. units
2 \(\frac{10 \sqrt{2}}{3}\) sq. units
3 \(\frac{15 \sqrt{2}}{3}\) sq. units
4 \(\frac{20 \sqrt{2}}{3}\) sq. units
Explanation:
(D) : Given the curves, \(y=4 x^{2}, y=\frac{x^{2}}{9} \text { and the line } y=2\) Let us draw the above. To find the point \(A\), solve \(y=4 x^{2}\) and \(y=2\) \(\therefore \quad 2=4 \mathrm{x}^{2}\) Or \(\quad x^{2}=\frac{1}{2} \quad\) Or \(\quad x=\frac{1}{\sqrt{2}}\) To find the point solve, \(y=\frac{x^{2}}{9} \text { and } y=2\) We get, \(\mathrm{x}^{2}=2 \times 9\) \(\therefore \quad \mathrm{x}=3 \sqrt{2}\) \(A_{1}=\int_{0}^{3 \sqrt{2}}\left(2-\frac{x^{2}}{9}\right) d x=2 \cdot x-\left.\frac{1}{9} \cdot \frac{x^{3}}{3}\right|_{0} ^{3 \sqrt{2}}\) \(=2 \cdot[3 \sqrt{2}]-\frac{1}{27} \cdot\left[(3 \sqrt{2})^{3}-0\right]\) \(=6 \sqrt{2}-\frac{1}{27} \times 27 \times 2 \sqrt{2}=6 \sqrt{2}-2 \sqrt{2}=4 \sqrt{2}\) \(A_{2}=\int_{0}^{1 / \sqrt{2}}\left(2-4 x^{2}\right) d x=2 \cdot \int_{0}^{1 / \sqrt{2}} x-4 \int_{0}^{1 / \sqrt{2}} x^{2} \cdot d x\) \(=2 \cdot \times \frac{1}{\sqrt{2}}-\left.\frac{4}{3} \cdot x^{3}\right|_{0} ^{1 / \sqrt{2}}=\sqrt{2}-\frac{4}{3} \times \frac{1}{2 \sqrt{2}}\) \(=\sqrt{2}-\frac{\sqrt{2}}{3}=\sqrt{2}\left(1-\frac{1}{3}\right)=\frac{2 \sqrt{2}}{3}\) \(\therefore\) Required area is \(\left(\mathrm{A}_{1}-\mathrm{A}_{2}\right) \times 2\). \(=\left(4 \sqrt{2}-\frac{2 \sqrt{2}}{3}\right) \times 2=\sqrt{2} .\left(4-\frac{2}{3}\right) \times 2\) \(=\frac{10}{3} \times 2 \sqrt{2}=\frac{20}{3} \sqrt{2}\) Square units.
WB JEE-2010
Application of the Integrals
87000
The area of the circle centred at \((-92,103)\) and passing through \((-95,99)\) is:
1 \(3 \pi\)
2 \(4 \pi\)
3 \(25 \pi\)
4 \(5 \pi\)
Explanation:
(C) : From figure- \(\text { OA }=\text { radius }\) \((x-a)^2+(y-b)^2=r^2\) \((-95+92)^2+(99-103)^2=r^2\) \((-3)^2+(-4)^2=r^2\) \(9+16=r^2\) \(r=5\) So, Area of circle \(=\pi \mathrm{r}^{2}=\pi(5)^{2}=25 \pi\)
AMU-2012
Application of the Integrals
87001
Area of the figure bounded by the parabola \(y^{2}+\) \(8 x=16\) and \(y^{2}-24 x=48\) is
1 \(\frac{11}{9}\) sq.unit
2 \(\frac{32}{3} \sqrt{6}\) sq.unit
3 \(\frac{16}{3}\) sq.unit
4 \(\frac{24}{5}\) sq.unit
Explanation:
(B) : Given, the curves \(\mathrm{y}^{2}+8 \mathrm{x}=16\) And \(y^{2}-24 x=48\), we get \(\mathrm{y}^{2}=-8 \mathrm{x}+16\) Then, \(\quad y^{2}=-8(x-2)\) And \(\quad y^{2}=24(x+2)\) Let, us plot these curves. Solving, these two curves, we get - \(-8 x+16=24 x+48\) \(32 x=-32 \Rightarrow x=-1\) \(\therefore \quad \mathrm{y}^{2}=24(-1+2)=24\) \(\therefore \quad \mathrm{y}= \pm 2 \sqrt{6}\) \(\therefore\) The required area \(\mathrm{A}\) \(=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[\left(2-\frac{y^{2}}{8}\right)-\left(\frac{y^{2}}{24}-2\right)\right] d y\) Therefore, \(y^{2}+8 x=16\) \(\therefore \quad 8 x =16-y^{2}\) \(x =2-\frac{y^{2}}{8} \tag{i}\) And,\(\quad y^{2}-24 x=48\) \(24 x=y^{2}-48\) \(x=\frac{y^{2}}{24}-2 \tag{ii}\) \(\therefore A=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[4-\frac{3 y^{2}}{24}-\frac{y^{2}}{24}\right] d y=[4 y]_{2 \sqrt{6}}^{-2 \sqrt{6}}-\frac{1}{6} \int_{2 \sqrt{6}}^{-2 \sqrt{6}} y^{2} \cdot d y\) \(=4 \cdot(-2 \times 2 \sqrt{6})-\frac{1}{6} \cdot\left[\frac{y^{3}}{3}\right]_{2 \sqrt{6}}^{-2 \sqrt{6}}\) \(=-16 \sqrt{6}-\frac{1}{18}\left(-(2 \sqrt{6})^{3}-(2 \sqrt{6})^{3}\right)\) \(=-16 \sqrt{6}+\frac{1}{18} \times 2 \times(2 \sqrt{6})^{3}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 2 \times 2 \times 2 \times 6 \sqrt{6}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 8 \times 6 \sqrt{6}=-16 \sqrt{6}+\frac{16 \sqrt{6}}{3}\) \(=16 \sqrt{6}\left(\frac{1}{3}-1\right)=16 \sqrt{6}\left(\frac{-2}{3}\right)=\frac{-32}{3} \sqrt{6}\) But the area is never negative, \(\therefore\) Required area is \(\frac{32}{3} \sqrt{6}\) square unit.
WB JEE-2022
Application of the Integrals
87002
The area (in sq units) of the region \(\left\{(x, y): y^{2} \geq\right.\) \(2 x\) and \(\left.x^{2}+y^{2} \leq 4 x, x \geq 0, y \geq 0\right\}\) is
1 \(\pi-\frac{4}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi-\frac{4 \sqrt{2}}{3}\)
4 \(\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}\)
Explanation:
(B) : We have, \(y^{2}=2 x \tag{i}\) Which is a parabola with vertex \((0,0)\) and its axis parallel to \(\mathrm{x}\) - axis. and another curve \(x^{2}+y^{2}=4 x\) Which is a circle with centre \((2,0)\) and radius is 2 . On putting equation in equation (ii), we get - \(x^{2}+2 x=4 x\) \(x^{2}=2 x\) \(x^{2}-2 x=0\) \(x(x-2)=0\) \(x=0 \text { or } x=2\) \(\therefore\) Putting the value of \(\mathrm{x}\) in equation (i) we get - \(y=0 \text { or } y= \pm 2\) Therefore, required area \(=\frac{\text { Area of circle }}{4}-\int_{0}^{2} \sqrt{2 \mathrm{x}} \mathrm{dx}\) \(=\frac{\pi(2)^{2}}{4}-\sqrt{2} \int_{0}^{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\pi-\sqrt{2}\left[\frac{\mathrm{x}^{3 / 2}}{3 / 2}\right]_{0}^{2}\) \(=\pi-\frac{2 \sqrt{2}}{3}[2 \sqrt{2}-0]=\left(\pi-\frac{8}{3}\right)\) square unit
86999
The area bounded by the parabolas \(y=4 x^{2}\), \(y=\frac{x^{2}}{9}\) and the line \(y=2\) is
1 \(\frac{5 \sqrt{2}}{3}\) sq. units
2 \(\frac{10 \sqrt{2}}{3}\) sq. units
3 \(\frac{15 \sqrt{2}}{3}\) sq. units
4 \(\frac{20 \sqrt{2}}{3}\) sq. units
Explanation:
(D) : Given the curves, \(y=4 x^{2}, y=\frac{x^{2}}{9} \text { and the line } y=2\) Let us draw the above. To find the point \(A\), solve \(y=4 x^{2}\) and \(y=2\) \(\therefore \quad 2=4 \mathrm{x}^{2}\) Or \(\quad x^{2}=\frac{1}{2} \quad\) Or \(\quad x=\frac{1}{\sqrt{2}}\) To find the point solve, \(y=\frac{x^{2}}{9} \text { and } y=2\) We get, \(\mathrm{x}^{2}=2 \times 9\) \(\therefore \quad \mathrm{x}=3 \sqrt{2}\) \(A_{1}=\int_{0}^{3 \sqrt{2}}\left(2-\frac{x^{2}}{9}\right) d x=2 \cdot x-\left.\frac{1}{9} \cdot \frac{x^{3}}{3}\right|_{0} ^{3 \sqrt{2}}\) \(=2 \cdot[3 \sqrt{2}]-\frac{1}{27} \cdot\left[(3 \sqrt{2})^{3}-0\right]\) \(=6 \sqrt{2}-\frac{1}{27} \times 27 \times 2 \sqrt{2}=6 \sqrt{2}-2 \sqrt{2}=4 \sqrt{2}\) \(A_{2}=\int_{0}^{1 / \sqrt{2}}\left(2-4 x^{2}\right) d x=2 \cdot \int_{0}^{1 / \sqrt{2}} x-4 \int_{0}^{1 / \sqrt{2}} x^{2} \cdot d x\) \(=2 \cdot \times \frac{1}{\sqrt{2}}-\left.\frac{4}{3} \cdot x^{3}\right|_{0} ^{1 / \sqrt{2}}=\sqrt{2}-\frac{4}{3} \times \frac{1}{2 \sqrt{2}}\) \(=\sqrt{2}-\frac{\sqrt{2}}{3}=\sqrt{2}\left(1-\frac{1}{3}\right)=\frac{2 \sqrt{2}}{3}\) \(\therefore\) Required area is \(\left(\mathrm{A}_{1}-\mathrm{A}_{2}\right) \times 2\). \(=\left(4 \sqrt{2}-\frac{2 \sqrt{2}}{3}\right) \times 2=\sqrt{2} .\left(4-\frac{2}{3}\right) \times 2\) \(=\frac{10}{3} \times 2 \sqrt{2}=\frac{20}{3} \sqrt{2}\) Square units.
WB JEE-2010
Application of the Integrals
87000
The area of the circle centred at \((-92,103)\) and passing through \((-95,99)\) is:
1 \(3 \pi\)
2 \(4 \pi\)
3 \(25 \pi\)
4 \(5 \pi\)
Explanation:
(C) : From figure- \(\text { OA }=\text { radius }\) \((x-a)^2+(y-b)^2=r^2\) \((-95+92)^2+(99-103)^2=r^2\) \((-3)^2+(-4)^2=r^2\) \(9+16=r^2\) \(r=5\) So, Area of circle \(=\pi \mathrm{r}^{2}=\pi(5)^{2}=25 \pi\)
AMU-2012
Application of the Integrals
87001
Area of the figure bounded by the parabola \(y^{2}+\) \(8 x=16\) and \(y^{2}-24 x=48\) is
1 \(\frac{11}{9}\) sq.unit
2 \(\frac{32}{3} \sqrt{6}\) sq.unit
3 \(\frac{16}{3}\) sq.unit
4 \(\frac{24}{5}\) sq.unit
Explanation:
(B) : Given, the curves \(\mathrm{y}^{2}+8 \mathrm{x}=16\) And \(y^{2}-24 x=48\), we get \(\mathrm{y}^{2}=-8 \mathrm{x}+16\) Then, \(\quad y^{2}=-8(x-2)\) And \(\quad y^{2}=24(x+2)\) Let, us plot these curves. Solving, these two curves, we get - \(-8 x+16=24 x+48\) \(32 x=-32 \Rightarrow x=-1\) \(\therefore \quad \mathrm{y}^{2}=24(-1+2)=24\) \(\therefore \quad \mathrm{y}= \pm 2 \sqrt{6}\) \(\therefore\) The required area \(\mathrm{A}\) \(=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[\left(2-\frac{y^{2}}{8}\right)-\left(\frac{y^{2}}{24}-2\right)\right] d y\) Therefore, \(y^{2}+8 x=16\) \(\therefore \quad 8 x =16-y^{2}\) \(x =2-\frac{y^{2}}{8} \tag{i}\) And,\(\quad y^{2}-24 x=48\) \(24 x=y^{2}-48\) \(x=\frac{y^{2}}{24}-2 \tag{ii}\) \(\therefore A=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[4-\frac{3 y^{2}}{24}-\frac{y^{2}}{24}\right] d y=[4 y]_{2 \sqrt{6}}^{-2 \sqrt{6}}-\frac{1}{6} \int_{2 \sqrt{6}}^{-2 \sqrt{6}} y^{2} \cdot d y\) \(=4 \cdot(-2 \times 2 \sqrt{6})-\frac{1}{6} \cdot\left[\frac{y^{3}}{3}\right]_{2 \sqrt{6}}^{-2 \sqrt{6}}\) \(=-16 \sqrt{6}-\frac{1}{18}\left(-(2 \sqrt{6})^{3}-(2 \sqrt{6})^{3}\right)\) \(=-16 \sqrt{6}+\frac{1}{18} \times 2 \times(2 \sqrt{6})^{3}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 2 \times 2 \times 2 \times 6 \sqrt{6}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 8 \times 6 \sqrt{6}=-16 \sqrt{6}+\frac{16 \sqrt{6}}{3}\) \(=16 \sqrt{6}\left(\frac{1}{3}-1\right)=16 \sqrt{6}\left(\frac{-2}{3}\right)=\frac{-32}{3} \sqrt{6}\) But the area is never negative, \(\therefore\) Required area is \(\frac{32}{3} \sqrt{6}\) square unit.
WB JEE-2022
Application of the Integrals
87002
The area (in sq units) of the region \(\left\{(x, y): y^{2} \geq\right.\) \(2 x\) and \(\left.x^{2}+y^{2} \leq 4 x, x \geq 0, y \geq 0\right\}\) is
1 \(\pi-\frac{4}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi-\frac{4 \sqrt{2}}{3}\)
4 \(\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}\)
Explanation:
(B) : We have, \(y^{2}=2 x \tag{i}\) Which is a parabola with vertex \((0,0)\) and its axis parallel to \(\mathrm{x}\) - axis. and another curve \(x^{2}+y^{2}=4 x\) Which is a circle with centre \((2,0)\) and radius is 2 . On putting equation in equation (ii), we get - \(x^{2}+2 x=4 x\) \(x^{2}=2 x\) \(x^{2}-2 x=0\) \(x(x-2)=0\) \(x=0 \text { or } x=2\) \(\therefore\) Putting the value of \(\mathrm{x}\) in equation (i) we get - \(y=0 \text { or } y= \pm 2\) Therefore, required area \(=\frac{\text { Area of circle }}{4}-\int_{0}^{2} \sqrt{2 \mathrm{x}} \mathrm{dx}\) \(=\frac{\pi(2)^{2}}{4}-\sqrt{2} \int_{0}^{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\pi-\sqrt{2}\left[\frac{\mathrm{x}^{3 / 2}}{3 / 2}\right]_{0}^{2}\) \(=\pi-\frac{2 \sqrt{2}}{3}[2 \sqrt{2}-0]=\left(\pi-\frac{8}{3}\right)\) square unit
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
86999
The area bounded by the parabolas \(y=4 x^{2}\), \(y=\frac{x^{2}}{9}\) and the line \(y=2\) is
1 \(\frac{5 \sqrt{2}}{3}\) sq. units
2 \(\frac{10 \sqrt{2}}{3}\) sq. units
3 \(\frac{15 \sqrt{2}}{3}\) sq. units
4 \(\frac{20 \sqrt{2}}{3}\) sq. units
Explanation:
(D) : Given the curves, \(y=4 x^{2}, y=\frac{x^{2}}{9} \text { and the line } y=2\) Let us draw the above. To find the point \(A\), solve \(y=4 x^{2}\) and \(y=2\) \(\therefore \quad 2=4 \mathrm{x}^{2}\) Or \(\quad x^{2}=\frac{1}{2} \quad\) Or \(\quad x=\frac{1}{\sqrt{2}}\) To find the point solve, \(y=\frac{x^{2}}{9} \text { and } y=2\) We get, \(\mathrm{x}^{2}=2 \times 9\) \(\therefore \quad \mathrm{x}=3 \sqrt{2}\) \(A_{1}=\int_{0}^{3 \sqrt{2}}\left(2-\frac{x^{2}}{9}\right) d x=2 \cdot x-\left.\frac{1}{9} \cdot \frac{x^{3}}{3}\right|_{0} ^{3 \sqrt{2}}\) \(=2 \cdot[3 \sqrt{2}]-\frac{1}{27} \cdot\left[(3 \sqrt{2})^{3}-0\right]\) \(=6 \sqrt{2}-\frac{1}{27} \times 27 \times 2 \sqrt{2}=6 \sqrt{2}-2 \sqrt{2}=4 \sqrt{2}\) \(A_{2}=\int_{0}^{1 / \sqrt{2}}\left(2-4 x^{2}\right) d x=2 \cdot \int_{0}^{1 / \sqrt{2}} x-4 \int_{0}^{1 / \sqrt{2}} x^{2} \cdot d x\) \(=2 \cdot \times \frac{1}{\sqrt{2}}-\left.\frac{4}{3} \cdot x^{3}\right|_{0} ^{1 / \sqrt{2}}=\sqrt{2}-\frac{4}{3} \times \frac{1}{2 \sqrt{2}}\) \(=\sqrt{2}-\frac{\sqrt{2}}{3}=\sqrt{2}\left(1-\frac{1}{3}\right)=\frac{2 \sqrt{2}}{3}\) \(\therefore\) Required area is \(\left(\mathrm{A}_{1}-\mathrm{A}_{2}\right) \times 2\). \(=\left(4 \sqrt{2}-\frac{2 \sqrt{2}}{3}\right) \times 2=\sqrt{2} .\left(4-\frac{2}{3}\right) \times 2\) \(=\frac{10}{3} \times 2 \sqrt{2}=\frac{20}{3} \sqrt{2}\) Square units.
WB JEE-2010
Application of the Integrals
87000
The area of the circle centred at \((-92,103)\) and passing through \((-95,99)\) is:
1 \(3 \pi\)
2 \(4 \pi\)
3 \(25 \pi\)
4 \(5 \pi\)
Explanation:
(C) : From figure- \(\text { OA }=\text { radius }\) \((x-a)^2+(y-b)^2=r^2\) \((-95+92)^2+(99-103)^2=r^2\) \((-3)^2+(-4)^2=r^2\) \(9+16=r^2\) \(r=5\) So, Area of circle \(=\pi \mathrm{r}^{2}=\pi(5)^{2}=25 \pi\)
AMU-2012
Application of the Integrals
87001
Area of the figure bounded by the parabola \(y^{2}+\) \(8 x=16\) and \(y^{2}-24 x=48\) is
1 \(\frac{11}{9}\) sq.unit
2 \(\frac{32}{3} \sqrt{6}\) sq.unit
3 \(\frac{16}{3}\) sq.unit
4 \(\frac{24}{5}\) sq.unit
Explanation:
(B) : Given, the curves \(\mathrm{y}^{2}+8 \mathrm{x}=16\) And \(y^{2}-24 x=48\), we get \(\mathrm{y}^{2}=-8 \mathrm{x}+16\) Then, \(\quad y^{2}=-8(x-2)\) And \(\quad y^{2}=24(x+2)\) Let, us plot these curves. Solving, these two curves, we get - \(-8 x+16=24 x+48\) \(32 x=-32 \Rightarrow x=-1\) \(\therefore \quad \mathrm{y}^{2}=24(-1+2)=24\) \(\therefore \quad \mathrm{y}= \pm 2 \sqrt{6}\) \(\therefore\) The required area \(\mathrm{A}\) \(=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[\left(2-\frac{y^{2}}{8}\right)-\left(\frac{y^{2}}{24}-2\right)\right] d y\) Therefore, \(y^{2}+8 x=16\) \(\therefore \quad 8 x =16-y^{2}\) \(x =2-\frac{y^{2}}{8} \tag{i}\) And,\(\quad y^{2}-24 x=48\) \(24 x=y^{2}-48\) \(x=\frac{y^{2}}{24}-2 \tag{ii}\) \(\therefore A=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[4-\frac{3 y^{2}}{24}-\frac{y^{2}}{24}\right] d y=[4 y]_{2 \sqrt{6}}^{-2 \sqrt{6}}-\frac{1}{6} \int_{2 \sqrt{6}}^{-2 \sqrt{6}} y^{2} \cdot d y\) \(=4 \cdot(-2 \times 2 \sqrt{6})-\frac{1}{6} \cdot\left[\frac{y^{3}}{3}\right]_{2 \sqrt{6}}^{-2 \sqrt{6}}\) \(=-16 \sqrt{6}-\frac{1}{18}\left(-(2 \sqrt{6})^{3}-(2 \sqrt{6})^{3}\right)\) \(=-16 \sqrt{6}+\frac{1}{18} \times 2 \times(2 \sqrt{6})^{3}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 2 \times 2 \times 2 \times 6 \sqrt{6}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 8 \times 6 \sqrt{6}=-16 \sqrt{6}+\frac{16 \sqrt{6}}{3}\) \(=16 \sqrt{6}\left(\frac{1}{3}-1\right)=16 \sqrt{6}\left(\frac{-2}{3}\right)=\frac{-32}{3} \sqrt{6}\) But the area is never negative, \(\therefore\) Required area is \(\frac{32}{3} \sqrt{6}\) square unit.
WB JEE-2022
Application of the Integrals
87002
The area (in sq units) of the region \(\left\{(x, y): y^{2} \geq\right.\) \(2 x\) and \(\left.x^{2}+y^{2} \leq 4 x, x \geq 0, y \geq 0\right\}\) is
1 \(\pi-\frac{4}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi-\frac{4 \sqrt{2}}{3}\)
4 \(\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}\)
Explanation:
(B) : We have, \(y^{2}=2 x \tag{i}\) Which is a parabola with vertex \((0,0)\) and its axis parallel to \(\mathrm{x}\) - axis. and another curve \(x^{2}+y^{2}=4 x\) Which is a circle with centre \((2,0)\) and radius is 2 . On putting equation in equation (ii), we get - \(x^{2}+2 x=4 x\) \(x^{2}=2 x\) \(x^{2}-2 x=0\) \(x(x-2)=0\) \(x=0 \text { or } x=2\) \(\therefore\) Putting the value of \(\mathrm{x}\) in equation (i) we get - \(y=0 \text { or } y= \pm 2\) Therefore, required area \(=\frac{\text { Area of circle }}{4}-\int_{0}^{2} \sqrt{2 \mathrm{x}} \mathrm{dx}\) \(=\frac{\pi(2)^{2}}{4}-\sqrt{2} \int_{0}^{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\pi-\sqrt{2}\left[\frac{\mathrm{x}^{3 / 2}}{3 / 2}\right]_{0}^{2}\) \(=\pi-\frac{2 \sqrt{2}}{3}[2 \sqrt{2}-0]=\left(\pi-\frac{8}{3}\right)\) square unit
86999
The area bounded by the parabolas \(y=4 x^{2}\), \(y=\frac{x^{2}}{9}\) and the line \(y=2\) is
1 \(\frac{5 \sqrt{2}}{3}\) sq. units
2 \(\frac{10 \sqrt{2}}{3}\) sq. units
3 \(\frac{15 \sqrt{2}}{3}\) sq. units
4 \(\frac{20 \sqrt{2}}{3}\) sq. units
Explanation:
(D) : Given the curves, \(y=4 x^{2}, y=\frac{x^{2}}{9} \text { and the line } y=2\) Let us draw the above. To find the point \(A\), solve \(y=4 x^{2}\) and \(y=2\) \(\therefore \quad 2=4 \mathrm{x}^{2}\) Or \(\quad x^{2}=\frac{1}{2} \quad\) Or \(\quad x=\frac{1}{\sqrt{2}}\) To find the point solve, \(y=\frac{x^{2}}{9} \text { and } y=2\) We get, \(\mathrm{x}^{2}=2 \times 9\) \(\therefore \quad \mathrm{x}=3 \sqrt{2}\) \(A_{1}=\int_{0}^{3 \sqrt{2}}\left(2-\frac{x^{2}}{9}\right) d x=2 \cdot x-\left.\frac{1}{9} \cdot \frac{x^{3}}{3}\right|_{0} ^{3 \sqrt{2}}\) \(=2 \cdot[3 \sqrt{2}]-\frac{1}{27} \cdot\left[(3 \sqrt{2})^{3}-0\right]\) \(=6 \sqrt{2}-\frac{1}{27} \times 27 \times 2 \sqrt{2}=6 \sqrt{2}-2 \sqrt{2}=4 \sqrt{2}\) \(A_{2}=\int_{0}^{1 / \sqrt{2}}\left(2-4 x^{2}\right) d x=2 \cdot \int_{0}^{1 / \sqrt{2}} x-4 \int_{0}^{1 / \sqrt{2}} x^{2} \cdot d x\) \(=2 \cdot \times \frac{1}{\sqrt{2}}-\left.\frac{4}{3} \cdot x^{3}\right|_{0} ^{1 / \sqrt{2}}=\sqrt{2}-\frac{4}{3} \times \frac{1}{2 \sqrt{2}}\) \(=\sqrt{2}-\frac{\sqrt{2}}{3}=\sqrt{2}\left(1-\frac{1}{3}\right)=\frac{2 \sqrt{2}}{3}\) \(\therefore\) Required area is \(\left(\mathrm{A}_{1}-\mathrm{A}_{2}\right) \times 2\). \(=\left(4 \sqrt{2}-\frac{2 \sqrt{2}}{3}\right) \times 2=\sqrt{2} .\left(4-\frac{2}{3}\right) \times 2\) \(=\frac{10}{3} \times 2 \sqrt{2}=\frac{20}{3} \sqrt{2}\) Square units.
WB JEE-2010
Application of the Integrals
87000
The area of the circle centred at \((-92,103)\) and passing through \((-95,99)\) is:
1 \(3 \pi\)
2 \(4 \pi\)
3 \(25 \pi\)
4 \(5 \pi\)
Explanation:
(C) : From figure- \(\text { OA }=\text { radius }\) \((x-a)^2+(y-b)^2=r^2\) \((-95+92)^2+(99-103)^2=r^2\) \((-3)^2+(-4)^2=r^2\) \(9+16=r^2\) \(r=5\) So, Area of circle \(=\pi \mathrm{r}^{2}=\pi(5)^{2}=25 \pi\)
AMU-2012
Application of the Integrals
87001
Area of the figure bounded by the parabola \(y^{2}+\) \(8 x=16\) and \(y^{2}-24 x=48\) is
1 \(\frac{11}{9}\) sq.unit
2 \(\frac{32}{3} \sqrt{6}\) sq.unit
3 \(\frac{16}{3}\) sq.unit
4 \(\frac{24}{5}\) sq.unit
Explanation:
(B) : Given, the curves \(\mathrm{y}^{2}+8 \mathrm{x}=16\) And \(y^{2}-24 x=48\), we get \(\mathrm{y}^{2}=-8 \mathrm{x}+16\) Then, \(\quad y^{2}=-8(x-2)\) And \(\quad y^{2}=24(x+2)\) Let, us plot these curves. Solving, these two curves, we get - \(-8 x+16=24 x+48\) \(32 x=-32 \Rightarrow x=-1\) \(\therefore \quad \mathrm{y}^{2}=24(-1+2)=24\) \(\therefore \quad \mathrm{y}= \pm 2 \sqrt{6}\) \(\therefore\) The required area \(\mathrm{A}\) \(=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[\left(2-\frac{y^{2}}{8}\right)-\left(\frac{y^{2}}{24}-2\right)\right] d y\) Therefore, \(y^{2}+8 x=16\) \(\therefore \quad 8 x =16-y^{2}\) \(x =2-\frac{y^{2}}{8} \tag{i}\) And,\(\quad y^{2}-24 x=48\) \(24 x=y^{2}-48\) \(x=\frac{y^{2}}{24}-2 \tag{ii}\) \(\therefore A=\int_{2 \sqrt{6}}^{-2 \sqrt{6}}\left[4-\frac{3 y^{2}}{24}-\frac{y^{2}}{24}\right] d y=[4 y]_{2 \sqrt{6}}^{-2 \sqrt{6}}-\frac{1}{6} \int_{2 \sqrt{6}}^{-2 \sqrt{6}} y^{2} \cdot d y\) \(=4 \cdot(-2 \times 2 \sqrt{6})-\frac{1}{6} \cdot\left[\frac{y^{3}}{3}\right]_{2 \sqrt{6}}^{-2 \sqrt{6}}\) \(=-16 \sqrt{6}-\frac{1}{18}\left(-(2 \sqrt{6})^{3}-(2 \sqrt{6})^{3}\right)\) \(=-16 \sqrt{6}+\frac{1}{18} \times 2 \times(2 \sqrt{6})^{3}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 2 \times 2 \times 2 \times 6 \sqrt{6}\) \(=-16 \sqrt{6}+\frac{1}{9} \times 8 \times 6 \sqrt{6}=-16 \sqrt{6}+\frac{16 \sqrt{6}}{3}\) \(=16 \sqrt{6}\left(\frac{1}{3}-1\right)=16 \sqrt{6}\left(\frac{-2}{3}\right)=\frac{-32}{3} \sqrt{6}\) But the area is never negative, \(\therefore\) Required area is \(\frac{32}{3} \sqrt{6}\) square unit.
WB JEE-2022
Application of the Integrals
87002
The area (in sq units) of the region \(\left\{(x, y): y^{2} \geq\right.\) \(2 x\) and \(\left.x^{2}+y^{2} \leq 4 x, x \geq 0, y \geq 0\right\}\) is
1 \(\pi-\frac{4}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi-\frac{4 \sqrt{2}}{3}\)
4 \(\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}\)
Explanation:
(B) : We have, \(y^{2}=2 x \tag{i}\) Which is a parabola with vertex \((0,0)\) and its axis parallel to \(\mathrm{x}\) - axis. and another curve \(x^{2}+y^{2}=4 x\) Which is a circle with centre \((2,0)\) and radius is 2 . On putting equation in equation (ii), we get - \(x^{2}+2 x=4 x\) \(x^{2}=2 x\) \(x^{2}-2 x=0\) \(x(x-2)=0\) \(x=0 \text { or } x=2\) \(\therefore\) Putting the value of \(\mathrm{x}\) in equation (i) we get - \(y=0 \text { or } y= \pm 2\) Therefore, required area \(=\frac{\text { Area of circle }}{4}-\int_{0}^{2} \sqrt{2 \mathrm{x}} \mathrm{dx}\) \(=\frac{\pi(2)^{2}}{4}-\sqrt{2} \int_{0}^{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\pi-\sqrt{2}\left[\frac{\mathrm{x}^{3 / 2}}{3 / 2}\right]_{0}^{2}\) \(=\pi-\frac{2 \sqrt{2}}{3}[2 \sqrt{2}-0]=\left(\pi-\frac{8}{3}\right)\) square unit
