86990
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to the parabola at the point \((2,3)\) and the \(X\)-axis is
1 3
2 6
3 9
4 12
Explanation:
(C) : Given, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) or \(\quad y^{2}-4 y-x+5=0\) Equation of the tangent to the parabola at \((2,3)\), \(y-3=\frac{1}{2}(x-2)\) \(x-2 y+4=0\) \(x=2 y-4\) Now required area, \(A=\int_{0}^{3}(y-2)^{2}+1-(2 y-4) \cdot d y\) \(A=\int_{0}^{3}\left(y^{2}-6 y+9\right) \cdot d y \Rightarrow A=\int_{0}^{3}(y-3)^{2} \cdot d y\) \(\mathrm{A}=\int_{0}^{3}(3-\mathrm{y})^{2} \cdot \mathrm{dy} \quad(\) let \(3-\mathrm{y}=\mathrm{t})\) \(\mathrm{A}=\int_{0}^{3}(\mathrm{t})^{2} \cdot \mathrm{dt}\) \(\mathrm{A}=\left[\frac{\mathrm{t}^{3}}{3}\right]_{0}^{3}=\frac{3^{3}}{3} \Rightarrow \mathrm{A}=9\) square units.
[JCECE-2015]
Application of the Integrals
86992
The area of the curve \(x^{2}+y^{2}=r^{2}\) is
1 \(4 \pi \mathrm{a}^{2}\)
2 \(\pi r^{2}\)
3 \(\frac{\pi \mathrm{a}^{2}}{2}\)
4 \(2 \pi \mathrm{a}^{2}\)
Explanation:
(B) : Equation \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}\) Represent circle with centre \((0,0)\) \(x=\sqrt{r^{2}-y^{2}}\) Area of the whole circle \(=4\) (Area of first quadrant) Required area \(4 \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y=4\left[\frac{y}{2} \sqrt{r^{2}-y^{2}}+\frac{r^{2}}{2} \sin ^{-1} \frac{y}{r}\right]_{0}^{r}\) \(=4\left[0+\frac{r^{2}}{2} \sin ^{-1}(1)-0\right]=\pi r^{2} \text { sq unit }\)
AMU-2002
Application of the Integrals
86993
The area bounded by the two parabolas \(y^{2}=x\) and \(x^{2}=y\) is given by
1 1
2 \(\frac{2}{3}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{2}\)
Explanation:
(C) : We have \(\mathrm{y}^{2}=\mathrm{x}\) and \(x^{2}=y\) Solving equation (i) and (ii) we get- \(\mathrm{x}=1, \mathrm{y}=1\) The curve \(\mathrm{y}^{2}=\mathrm{x}\) and \(\mathrm{x}^{2}=\mathrm{y}\) Intersect at \((1,1)\) Hence, Required area \(\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx} \Rightarrow \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1}-\frac{1}{3}\left[\mathrm{x}^{3}\right]_{0}^{1}\) \(\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
AMU-2011
Application of the Integrals
86994
The area of the smaller region enclosed by the curve \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3 x}-4=0\) is equal to
1 \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
2 \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
3 \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
4 \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Explanation:
(C) : Given curves, \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) Point of intersections are \((0,2)\) and \((0,-2)\) Area \(=\frac{16}{3}\) \(2 \int_{0}^{2} \sqrt{16-y^{2}} d y-2 \int_{0}^{2} 2 \sqrt{3} y-2 \int_{0}^{2} \frac{y^{2}}{8} d y+2 \int_{0}^{2} \frac{4}{8} d y\) \(=2\left[\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1} \frac{y}{4}\right]_{0}^{2}-\left.4 \sqrt{3} y\right|_{0} ^{2}-\left.\frac{1}{4} \frac{y^{3}}{3}\right|_{0} ^{2}+\left.y\right|_{0} ^{2}\) \(=2\left[\sqrt{16-4}+8 \sin ^{-1} \frac{1}{2}-0-8 \cdot \sin ^{-1} 0\right]-4 \sqrt{3} \cdot(2-0)-\frac{1}{12}(8-0)+(2-0)\) \(=2\left[\sqrt{12}+8 \cdot \frac{\pi}{6}\right]-8 \sqrt{3}-\frac{8}{12}+2\) \(=4 \sqrt{3}+\frac{8}{3} \pi-8 \sqrt{3}-\frac{2}{3}+2\) \(=\frac{8}{3} \pi-4 \sqrt{3}+\frac{4}{3}=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\) Area \(=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\)
JEE Main-2022-28.07.2022
Application of the Integrals
86995
Area of the region \(\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq\right.\) \(2 y\}\) is
1 \(2 \pi-\frac{16}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi+\frac{8}{3}\)
4 \(2 \pi+\frac{16}{3}\)
Explanation:
(A) : \(x^{2}+(y-2)^{2} \leq 2^{2}\) and \(x^{2} \geq 2 y\) Solving circle and parabola simultaneously: \(2 y+y^{2}-4 y+4=4\) \(y^{2}-2 y=0\) \(y=0,2\) Put \(\mathrm{y}=2\) in \(\mathrm{x}^{2}=2 \mathrm{y} \rightarrow \mathrm{x}= \pm 2\) \(\Rightarrow \quad(2,2)\) and \((-2,2)\) \(=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^{2}=4-\pi\) Required area \(=2\left[\int_{0}^{2} \frac{x^{2}}{2} d x-(4-\pi)\right]\) \(=2\left[\left.\frac{x^{3}}{6}\right|_{0} ^{2}-4+\pi\right]=2\left[\frac{4}{3}+\pi-4\right]=2\left[\pi-\frac{8}{3}\right]=2 \pi-\frac{16}{3}\)
86990
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to the parabola at the point \((2,3)\) and the \(X\)-axis is
1 3
2 6
3 9
4 12
Explanation:
(C) : Given, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) or \(\quad y^{2}-4 y-x+5=0\) Equation of the tangent to the parabola at \((2,3)\), \(y-3=\frac{1}{2}(x-2)\) \(x-2 y+4=0\) \(x=2 y-4\) Now required area, \(A=\int_{0}^{3}(y-2)^{2}+1-(2 y-4) \cdot d y\) \(A=\int_{0}^{3}\left(y^{2}-6 y+9\right) \cdot d y \Rightarrow A=\int_{0}^{3}(y-3)^{2} \cdot d y\) \(\mathrm{A}=\int_{0}^{3}(3-\mathrm{y})^{2} \cdot \mathrm{dy} \quad(\) let \(3-\mathrm{y}=\mathrm{t})\) \(\mathrm{A}=\int_{0}^{3}(\mathrm{t})^{2} \cdot \mathrm{dt}\) \(\mathrm{A}=\left[\frac{\mathrm{t}^{3}}{3}\right]_{0}^{3}=\frac{3^{3}}{3} \Rightarrow \mathrm{A}=9\) square units.
[JCECE-2015]
Application of the Integrals
86992
The area of the curve \(x^{2}+y^{2}=r^{2}\) is
1 \(4 \pi \mathrm{a}^{2}\)
2 \(\pi r^{2}\)
3 \(\frac{\pi \mathrm{a}^{2}}{2}\)
4 \(2 \pi \mathrm{a}^{2}\)
Explanation:
(B) : Equation \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}\) Represent circle with centre \((0,0)\) \(x=\sqrt{r^{2}-y^{2}}\) Area of the whole circle \(=4\) (Area of first quadrant) Required area \(4 \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y=4\left[\frac{y}{2} \sqrt{r^{2}-y^{2}}+\frac{r^{2}}{2} \sin ^{-1} \frac{y}{r}\right]_{0}^{r}\) \(=4\left[0+\frac{r^{2}}{2} \sin ^{-1}(1)-0\right]=\pi r^{2} \text { sq unit }\)
AMU-2002
Application of the Integrals
86993
The area bounded by the two parabolas \(y^{2}=x\) and \(x^{2}=y\) is given by
1 1
2 \(\frac{2}{3}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{2}\)
Explanation:
(C) : We have \(\mathrm{y}^{2}=\mathrm{x}\) and \(x^{2}=y\) Solving equation (i) and (ii) we get- \(\mathrm{x}=1, \mathrm{y}=1\) The curve \(\mathrm{y}^{2}=\mathrm{x}\) and \(\mathrm{x}^{2}=\mathrm{y}\) Intersect at \((1,1)\) Hence, Required area \(\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx} \Rightarrow \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1}-\frac{1}{3}\left[\mathrm{x}^{3}\right]_{0}^{1}\) \(\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
AMU-2011
Application of the Integrals
86994
The area of the smaller region enclosed by the curve \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3 x}-4=0\) is equal to
1 \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
2 \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
3 \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
4 \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Explanation:
(C) : Given curves, \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) Point of intersections are \((0,2)\) and \((0,-2)\) Area \(=\frac{16}{3}\) \(2 \int_{0}^{2} \sqrt{16-y^{2}} d y-2 \int_{0}^{2} 2 \sqrt{3} y-2 \int_{0}^{2} \frac{y^{2}}{8} d y+2 \int_{0}^{2} \frac{4}{8} d y\) \(=2\left[\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1} \frac{y}{4}\right]_{0}^{2}-\left.4 \sqrt{3} y\right|_{0} ^{2}-\left.\frac{1}{4} \frac{y^{3}}{3}\right|_{0} ^{2}+\left.y\right|_{0} ^{2}\) \(=2\left[\sqrt{16-4}+8 \sin ^{-1} \frac{1}{2}-0-8 \cdot \sin ^{-1} 0\right]-4 \sqrt{3} \cdot(2-0)-\frac{1}{12}(8-0)+(2-0)\) \(=2\left[\sqrt{12}+8 \cdot \frac{\pi}{6}\right]-8 \sqrt{3}-\frac{8}{12}+2\) \(=4 \sqrt{3}+\frac{8}{3} \pi-8 \sqrt{3}-\frac{2}{3}+2\) \(=\frac{8}{3} \pi-4 \sqrt{3}+\frac{4}{3}=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\) Area \(=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\)
JEE Main-2022-28.07.2022
Application of the Integrals
86995
Area of the region \(\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq\right.\) \(2 y\}\) is
1 \(2 \pi-\frac{16}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi+\frac{8}{3}\)
4 \(2 \pi+\frac{16}{3}\)
Explanation:
(A) : \(x^{2}+(y-2)^{2} \leq 2^{2}\) and \(x^{2} \geq 2 y\) Solving circle and parabola simultaneously: \(2 y+y^{2}-4 y+4=4\) \(y^{2}-2 y=0\) \(y=0,2\) Put \(\mathrm{y}=2\) in \(\mathrm{x}^{2}=2 \mathrm{y} \rightarrow \mathrm{x}= \pm 2\) \(\Rightarrow \quad(2,2)\) and \((-2,2)\) \(=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^{2}=4-\pi\) Required area \(=2\left[\int_{0}^{2} \frac{x^{2}}{2} d x-(4-\pi)\right]\) \(=2\left[\left.\frac{x^{3}}{6}\right|_{0} ^{2}-4+\pi\right]=2\left[\frac{4}{3}+\pi-4\right]=2\left[\pi-\frac{8}{3}\right]=2 \pi-\frac{16}{3}\)
86990
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to the parabola at the point \((2,3)\) and the \(X\)-axis is
1 3
2 6
3 9
4 12
Explanation:
(C) : Given, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) or \(\quad y^{2}-4 y-x+5=0\) Equation of the tangent to the parabola at \((2,3)\), \(y-3=\frac{1}{2}(x-2)\) \(x-2 y+4=0\) \(x=2 y-4\) Now required area, \(A=\int_{0}^{3}(y-2)^{2}+1-(2 y-4) \cdot d y\) \(A=\int_{0}^{3}\left(y^{2}-6 y+9\right) \cdot d y \Rightarrow A=\int_{0}^{3}(y-3)^{2} \cdot d y\) \(\mathrm{A}=\int_{0}^{3}(3-\mathrm{y})^{2} \cdot \mathrm{dy} \quad(\) let \(3-\mathrm{y}=\mathrm{t})\) \(\mathrm{A}=\int_{0}^{3}(\mathrm{t})^{2} \cdot \mathrm{dt}\) \(\mathrm{A}=\left[\frac{\mathrm{t}^{3}}{3}\right]_{0}^{3}=\frac{3^{3}}{3} \Rightarrow \mathrm{A}=9\) square units.
[JCECE-2015]
Application of the Integrals
86992
The area of the curve \(x^{2}+y^{2}=r^{2}\) is
1 \(4 \pi \mathrm{a}^{2}\)
2 \(\pi r^{2}\)
3 \(\frac{\pi \mathrm{a}^{2}}{2}\)
4 \(2 \pi \mathrm{a}^{2}\)
Explanation:
(B) : Equation \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}\) Represent circle with centre \((0,0)\) \(x=\sqrt{r^{2}-y^{2}}\) Area of the whole circle \(=4\) (Area of first quadrant) Required area \(4 \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y=4\left[\frac{y}{2} \sqrt{r^{2}-y^{2}}+\frac{r^{2}}{2} \sin ^{-1} \frac{y}{r}\right]_{0}^{r}\) \(=4\left[0+\frac{r^{2}}{2} \sin ^{-1}(1)-0\right]=\pi r^{2} \text { sq unit }\)
AMU-2002
Application of the Integrals
86993
The area bounded by the two parabolas \(y^{2}=x\) and \(x^{2}=y\) is given by
1 1
2 \(\frac{2}{3}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{2}\)
Explanation:
(C) : We have \(\mathrm{y}^{2}=\mathrm{x}\) and \(x^{2}=y\) Solving equation (i) and (ii) we get- \(\mathrm{x}=1, \mathrm{y}=1\) The curve \(\mathrm{y}^{2}=\mathrm{x}\) and \(\mathrm{x}^{2}=\mathrm{y}\) Intersect at \((1,1)\) Hence, Required area \(\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx} \Rightarrow \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1}-\frac{1}{3}\left[\mathrm{x}^{3}\right]_{0}^{1}\) \(\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
AMU-2011
Application of the Integrals
86994
The area of the smaller region enclosed by the curve \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3 x}-4=0\) is equal to
1 \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
2 \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
3 \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
4 \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Explanation:
(C) : Given curves, \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) Point of intersections are \((0,2)\) and \((0,-2)\) Area \(=\frac{16}{3}\) \(2 \int_{0}^{2} \sqrt{16-y^{2}} d y-2 \int_{0}^{2} 2 \sqrt{3} y-2 \int_{0}^{2} \frac{y^{2}}{8} d y+2 \int_{0}^{2} \frac{4}{8} d y\) \(=2\left[\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1} \frac{y}{4}\right]_{0}^{2}-\left.4 \sqrt{3} y\right|_{0} ^{2}-\left.\frac{1}{4} \frac{y^{3}}{3}\right|_{0} ^{2}+\left.y\right|_{0} ^{2}\) \(=2\left[\sqrt{16-4}+8 \sin ^{-1} \frac{1}{2}-0-8 \cdot \sin ^{-1} 0\right]-4 \sqrt{3} \cdot(2-0)-\frac{1}{12}(8-0)+(2-0)\) \(=2\left[\sqrt{12}+8 \cdot \frac{\pi}{6}\right]-8 \sqrt{3}-\frac{8}{12}+2\) \(=4 \sqrt{3}+\frac{8}{3} \pi-8 \sqrt{3}-\frac{2}{3}+2\) \(=\frac{8}{3} \pi-4 \sqrt{3}+\frac{4}{3}=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\) Area \(=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\)
JEE Main-2022-28.07.2022
Application of the Integrals
86995
Area of the region \(\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq\right.\) \(2 y\}\) is
1 \(2 \pi-\frac{16}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi+\frac{8}{3}\)
4 \(2 \pi+\frac{16}{3}\)
Explanation:
(A) : \(x^{2}+(y-2)^{2} \leq 2^{2}\) and \(x^{2} \geq 2 y\) Solving circle and parabola simultaneously: \(2 y+y^{2}-4 y+4=4\) \(y^{2}-2 y=0\) \(y=0,2\) Put \(\mathrm{y}=2\) in \(\mathrm{x}^{2}=2 \mathrm{y} \rightarrow \mathrm{x}= \pm 2\) \(\Rightarrow \quad(2,2)\) and \((-2,2)\) \(=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^{2}=4-\pi\) Required area \(=2\left[\int_{0}^{2} \frac{x^{2}}{2} d x-(4-\pi)\right]\) \(=2\left[\left.\frac{x^{3}}{6}\right|_{0} ^{2}-4+\pi\right]=2\left[\frac{4}{3}+\pi-4\right]=2\left[\pi-\frac{8}{3}\right]=2 \pi-\frac{16}{3}\)
86990
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to the parabola at the point \((2,3)\) and the \(X\)-axis is
1 3
2 6
3 9
4 12
Explanation:
(C) : Given, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) or \(\quad y^{2}-4 y-x+5=0\) Equation of the tangent to the parabola at \((2,3)\), \(y-3=\frac{1}{2}(x-2)\) \(x-2 y+4=0\) \(x=2 y-4\) Now required area, \(A=\int_{0}^{3}(y-2)^{2}+1-(2 y-4) \cdot d y\) \(A=\int_{0}^{3}\left(y^{2}-6 y+9\right) \cdot d y \Rightarrow A=\int_{0}^{3}(y-3)^{2} \cdot d y\) \(\mathrm{A}=\int_{0}^{3}(3-\mathrm{y})^{2} \cdot \mathrm{dy} \quad(\) let \(3-\mathrm{y}=\mathrm{t})\) \(\mathrm{A}=\int_{0}^{3}(\mathrm{t})^{2} \cdot \mathrm{dt}\) \(\mathrm{A}=\left[\frac{\mathrm{t}^{3}}{3}\right]_{0}^{3}=\frac{3^{3}}{3} \Rightarrow \mathrm{A}=9\) square units.
[JCECE-2015]
Application of the Integrals
86992
The area of the curve \(x^{2}+y^{2}=r^{2}\) is
1 \(4 \pi \mathrm{a}^{2}\)
2 \(\pi r^{2}\)
3 \(\frac{\pi \mathrm{a}^{2}}{2}\)
4 \(2 \pi \mathrm{a}^{2}\)
Explanation:
(B) : Equation \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}\) Represent circle with centre \((0,0)\) \(x=\sqrt{r^{2}-y^{2}}\) Area of the whole circle \(=4\) (Area of first quadrant) Required area \(4 \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y=4\left[\frac{y}{2} \sqrt{r^{2}-y^{2}}+\frac{r^{2}}{2} \sin ^{-1} \frac{y}{r}\right]_{0}^{r}\) \(=4\left[0+\frac{r^{2}}{2} \sin ^{-1}(1)-0\right]=\pi r^{2} \text { sq unit }\)
AMU-2002
Application of the Integrals
86993
The area bounded by the two parabolas \(y^{2}=x\) and \(x^{2}=y\) is given by
1 1
2 \(\frac{2}{3}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{2}\)
Explanation:
(C) : We have \(\mathrm{y}^{2}=\mathrm{x}\) and \(x^{2}=y\) Solving equation (i) and (ii) we get- \(\mathrm{x}=1, \mathrm{y}=1\) The curve \(\mathrm{y}^{2}=\mathrm{x}\) and \(\mathrm{x}^{2}=\mathrm{y}\) Intersect at \((1,1)\) Hence, Required area \(\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx} \Rightarrow \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1}-\frac{1}{3}\left[\mathrm{x}^{3}\right]_{0}^{1}\) \(\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
AMU-2011
Application of the Integrals
86994
The area of the smaller region enclosed by the curve \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3 x}-4=0\) is equal to
1 \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
2 \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
3 \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
4 \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Explanation:
(C) : Given curves, \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) Point of intersections are \((0,2)\) and \((0,-2)\) Area \(=\frac{16}{3}\) \(2 \int_{0}^{2} \sqrt{16-y^{2}} d y-2 \int_{0}^{2} 2 \sqrt{3} y-2 \int_{0}^{2} \frac{y^{2}}{8} d y+2 \int_{0}^{2} \frac{4}{8} d y\) \(=2\left[\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1} \frac{y}{4}\right]_{0}^{2}-\left.4 \sqrt{3} y\right|_{0} ^{2}-\left.\frac{1}{4} \frac{y^{3}}{3}\right|_{0} ^{2}+\left.y\right|_{0} ^{2}\) \(=2\left[\sqrt{16-4}+8 \sin ^{-1} \frac{1}{2}-0-8 \cdot \sin ^{-1} 0\right]-4 \sqrt{3} \cdot(2-0)-\frac{1}{12}(8-0)+(2-0)\) \(=2\left[\sqrt{12}+8 \cdot \frac{\pi}{6}\right]-8 \sqrt{3}-\frac{8}{12}+2\) \(=4 \sqrt{3}+\frac{8}{3} \pi-8 \sqrt{3}-\frac{2}{3}+2\) \(=\frac{8}{3} \pi-4 \sqrt{3}+\frac{4}{3}=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\) Area \(=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\)
JEE Main-2022-28.07.2022
Application of the Integrals
86995
Area of the region \(\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq\right.\) \(2 y\}\) is
1 \(2 \pi-\frac{16}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi+\frac{8}{3}\)
4 \(2 \pi+\frac{16}{3}\)
Explanation:
(A) : \(x^{2}+(y-2)^{2} \leq 2^{2}\) and \(x^{2} \geq 2 y\) Solving circle and parabola simultaneously: \(2 y+y^{2}-4 y+4=4\) \(y^{2}-2 y=0\) \(y=0,2\) Put \(\mathrm{y}=2\) in \(\mathrm{x}^{2}=2 \mathrm{y} \rightarrow \mathrm{x}= \pm 2\) \(\Rightarrow \quad(2,2)\) and \((-2,2)\) \(=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^{2}=4-\pi\) Required area \(=2\left[\int_{0}^{2} \frac{x^{2}}{2} d x-(4-\pi)\right]\) \(=2\left[\left.\frac{x^{3}}{6}\right|_{0} ^{2}-4+\pi\right]=2\left[\frac{4}{3}+\pi-4\right]=2\left[\pi-\frac{8}{3}\right]=2 \pi-\frac{16}{3}\)
86990
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to the parabola at the point \((2,3)\) and the \(X\)-axis is
1 3
2 6
3 9
4 12
Explanation:
(C) : Given, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) or \(\quad y^{2}-4 y-x+5=0\) Equation of the tangent to the parabola at \((2,3)\), \(y-3=\frac{1}{2}(x-2)\) \(x-2 y+4=0\) \(x=2 y-4\) Now required area, \(A=\int_{0}^{3}(y-2)^{2}+1-(2 y-4) \cdot d y\) \(A=\int_{0}^{3}\left(y^{2}-6 y+9\right) \cdot d y \Rightarrow A=\int_{0}^{3}(y-3)^{2} \cdot d y\) \(\mathrm{A}=\int_{0}^{3}(3-\mathrm{y})^{2} \cdot \mathrm{dy} \quad(\) let \(3-\mathrm{y}=\mathrm{t})\) \(\mathrm{A}=\int_{0}^{3}(\mathrm{t})^{2} \cdot \mathrm{dt}\) \(\mathrm{A}=\left[\frac{\mathrm{t}^{3}}{3}\right]_{0}^{3}=\frac{3^{3}}{3} \Rightarrow \mathrm{A}=9\) square units.
[JCECE-2015]
Application of the Integrals
86992
The area of the curve \(x^{2}+y^{2}=r^{2}\) is
1 \(4 \pi \mathrm{a}^{2}\)
2 \(\pi r^{2}\)
3 \(\frac{\pi \mathrm{a}^{2}}{2}\)
4 \(2 \pi \mathrm{a}^{2}\)
Explanation:
(B) : Equation \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}\) Represent circle with centre \((0,0)\) \(x=\sqrt{r^{2}-y^{2}}\) Area of the whole circle \(=4\) (Area of first quadrant) Required area \(4 \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y=4\left[\frac{y}{2} \sqrt{r^{2}-y^{2}}+\frac{r^{2}}{2} \sin ^{-1} \frac{y}{r}\right]_{0}^{r}\) \(=4\left[0+\frac{r^{2}}{2} \sin ^{-1}(1)-0\right]=\pi r^{2} \text { sq unit }\)
AMU-2002
Application of the Integrals
86993
The area bounded by the two parabolas \(y^{2}=x\) and \(x^{2}=y\) is given by
1 1
2 \(\frac{2}{3}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{2}\)
Explanation:
(C) : We have \(\mathrm{y}^{2}=\mathrm{x}\) and \(x^{2}=y\) Solving equation (i) and (ii) we get- \(\mathrm{x}=1, \mathrm{y}=1\) The curve \(\mathrm{y}^{2}=\mathrm{x}\) and \(\mathrm{x}^{2}=\mathrm{y}\) Intersect at \((1,1)\) Hence, Required area \(\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx} \Rightarrow \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1}-\frac{1}{3}\left[\mathrm{x}^{3}\right]_{0}^{1}\) \(\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
AMU-2011
Application of the Integrals
86994
The area of the smaller region enclosed by the curve \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3 x}-4=0\) is equal to
1 \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
2 \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
3 \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
4 \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Explanation:
(C) : Given curves, \(y^{2}=8 x+4\) and \(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) Point of intersections are \((0,2)\) and \((0,-2)\) Area \(=\frac{16}{3}\) \(2 \int_{0}^{2} \sqrt{16-y^{2}} d y-2 \int_{0}^{2} 2 \sqrt{3} y-2 \int_{0}^{2} \frac{y^{2}}{8} d y+2 \int_{0}^{2} \frac{4}{8} d y\) \(=2\left[\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1} \frac{y}{4}\right]_{0}^{2}-\left.4 \sqrt{3} y\right|_{0} ^{2}-\left.\frac{1}{4} \frac{y^{3}}{3}\right|_{0} ^{2}+\left.y\right|_{0} ^{2}\) \(=2\left[\sqrt{16-4}+8 \sin ^{-1} \frac{1}{2}-0-8 \cdot \sin ^{-1} 0\right]-4 \sqrt{3} \cdot(2-0)-\frac{1}{12}(8-0)+(2-0)\) \(=2\left[\sqrt{12}+8 \cdot \frac{\pi}{6}\right]-8 \sqrt{3}-\frac{8}{12}+2\) \(=4 \sqrt{3}+\frac{8}{3} \pi-8 \sqrt{3}-\frac{2}{3}+2\) \(=\frac{8}{3} \pi-4 \sqrt{3}+\frac{4}{3}=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\) Area \(=\frac{1}{3}[8 \pi+4-12 \sqrt{3}]\)
JEE Main-2022-28.07.2022
Application of the Integrals
86995
Area of the region \(\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq\right.\) \(2 y\}\) is
1 \(2 \pi-\frac{16}{3}\)
2 \(\pi-\frac{8}{3}\)
3 \(\pi+\frac{8}{3}\)
4 \(2 \pi+\frac{16}{3}\)
Explanation:
(A) : \(x^{2}+(y-2)^{2} \leq 2^{2}\) and \(x^{2} \geq 2 y\) Solving circle and parabola simultaneously: \(2 y+y^{2}-4 y+4=4\) \(y^{2}-2 y=0\) \(y=0,2\) Put \(\mathrm{y}=2\) in \(\mathrm{x}^{2}=2 \mathrm{y} \rightarrow \mathrm{x}= \pm 2\) \(\Rightarrow \quad(2,2)\) and \((-2,2)\) \(=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^{2}=4-\pi\) Required area \(=2\left[\int_{0}^{2} \frac{x^{2}}{2} d x-(4-\pi)\right]\) \(=2\left[\left.\frac{x^{3}}{6}\right|_{0} ^{2}-4+\pi\right]=2\left[\frac{4}{3}+\pi-4\right]=2\left[\pi-\frac{8}{3}\right]=2 \pi-\frac{16}{3}\)
