NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86986
The area in the first quadrant between \(x^{2}+y^{2}=\pi^{2}\) and \(y=\sin x\) is
1 \(\frac{\pi^{3}-8}{4}\)
2 \(\frac{\pi^{3}}{4}\)
3 \(\frac{\pi^{3}-16}{4}\)
4 \(\frac{\pi^{3}-8}{2}\)
Explanation:
(A) : \(\mathrm{x}^{2}+\mathrm{y}^{2}=\pi^{2}\) is a circle of radius \(\pi\) and centre at origin. Required area \(=\) Area of circle \(\left(1^{\text {st }}\right.\) quadrant \()\) \(-\int_{0}^{\pi} \sin \mathrm{x} d \mathrm{x}\) \(=\frac{\pi \cdot \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)\) \(=\frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2 \Rightarrow \mathrm{A}=\frac{\pi^{3}-8}{4}\)
VITEEE-2014
Application of the Integrals
86987
A closed figure \(S\) is bounded by the hyperbola \(\mathbf{x}^{2}-\mathbf{y}^{2}=\mathbf{a}^{2}\) and the straight line \(x=a+h ;(h>\) \(0, a>0)\). This closed figure is rotated about the \(\mathrm{X}\)-axis. Then, the volume of the solid of revolution is
86986
The area in the first quadrant between \(x^{2}+y^{2}=\pi^{2}\) and \(y=\sin x\) is
1 \(\frac{\pi^{3}-8}{4}\)
2 \(\frac{\pi^{3}}{4}\)
3 \(\frac{\pi^{3}-16}{4}\)
4 \(\frac{\pi^{3}-8}{2}\)
Explanation:
(A) : \(\mathrm{x}^{2}+\mathrm{y}^{2}=\pi^{2}\) is a circle of radius \(\pi\) and centre at origin. Required area \(=\) Area of circle \(\left(1^{\text {st }}\right.\) quadrant \()\) \(-\int_{0}^{\pi} \sin \mathrm{x} d \mathrm{x}\) \(=\frac{\pi \cdot \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)\) \(=\frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2 \Rightarrow \mathrm{A}=\frac{\pi^{3}-8}{4}\)
VITEEE-2014
Application of the Integrals
86987
A closed figure \(S\) is bounded by the hyperbola \(\mathbf{x}^{2}-\mathbf{y}^{2}=\mathbf{a}^{2}\) and the straight line \(x=a+h ;(h>\) \(0, a>0)\). This closed figure is rotated about the \(\mathrm{X}\)-axis. Then, the volume of the solid of revolution is
86986
The area in the first quadrant between \(x^{2}+y^{2}=\pi^{2}\) and \(y=\sin x\) is
1 \(\frac{\pi^{3}-8}{4}\)
2 \(\frac{\pi^{3}}{4}\)
3 \(\frac{\pi^{3}-16}{4}\)
4 \(\frac{\pi^{3}-8}{2}\)
Explanation:
(A) : \(\mathrm{x}^{2}+\mathrm{y}^{2}=\pi^{2}\) is a circle of radius \(\pi\) and centre at origin. Required area \(=\) Area of circle \(\left(1^{\text {st }}\right.\) quadrant \()\) \(-\int_{0}^{\pi} \sin \mathrm{x} d \mathrm{x}\) \(=\frac{\pi \cdot \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)\) \(=\frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2 \Rightarrow \mathrm{A}=\frac{\pi^{3}-8}{4}\)
VITEEE-2014
Application of the Integrals
86987
A closed figure \(S\) is bounded by the hyperbola \(\mathbf{x}^{2}-\mathbf{y}^{2}=\mathbf{a}^{2}\) and the straight line \(x=a+h ;(h>\) \(0, a>0)\). This closed figure is rotated about the \(\mathrm{X}\)-axis. Then, the volume of the solid of revolution is
86986
The area in the first quadrant between \(x^{2}+y^{2}=\pi^{2}\) and \(y=\sin x\) is
1 \(\frac{\pi^{3}-8}{4}\)
2 \(\frac{\pi^{3}}{4}\)
3 \(\frac{\pi^{3}-16}{4}\)
4 \(\frac{\pi^{3}-8}{2}\)
Explanation:
(A) : \(\mathrm{x}^{2}+\mathrm{y}^{2}=\pi^{2}\) is a circle of radius \(\pi\) and centre at origin. Required area \(=\) Area of circle \(\left(1^{\text {st }}\right.\) quadrant \()\) \(-\int_{0}^{\pi} \sin \mathrm{x} d \mathrm{x}\) \(=\frac{\pi \cdot \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)\) \(=\frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2 \Rightarrow \mathrm{A}=\frac{\pi^{3}-8}{4}\)
VITEEE-2014
Application of the Integrals
86987
A closed figure \(S\) is bounded by the hyperbola \(\mathbf{x}^{2}-\mathbf{y}^{2}=\mathbf{a}^{2}\) and the straight line \(x=a+h ;(h>\) \(0, a>0)\). This closed figure is rotated about the \(\mathrm{X}\)-axis. Then, the volume of the solid of revolution is
