GATE Mechanical 2014 Set 1 | Question: 26

Arjun asked in Calculus Feb 19, 2017 recategorized Mar 4, 2021 by Lakshman Patel RJIT

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The integral $\oint_{c}^{ } (ydx-xdy)$ is evaluated along the circle $x^2+y^2=\frac{1}{4}$ traversed in counter clockwise direction. The integral is equal to

$0$
$\frac{-\pi }{4}$
$\frac{-\pi }{2}$
$\frac{\pi }{4}$

gateme-2014-set1
calculus
definite-integrals
area-under-curve

Arjun asked in Calculus Feb 19, 2017 recategorized Mar 4, 2021 by Lakshman Patel RJIT

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Andrijana3306 asked in Calculus Mar 20, 2018

GATE ME 2012 | Question: 11
The area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x-y$ plane is $1/6$ $1/4$ $1/3$ $1/2$

Andrijana3306 asked in Calculus Mar 20, 2018

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definite-integrals
area-under-curve

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Arjun asked in Calculus Feb 19, 2017

GATE Mechanical 2014 Set 4 | Question: 26
If $z$ is a complex variable, the value of $\displaystyle{} \int_{5}^{3i}\dfrac{dz}{z}$ is $− 0.511−1.57i$ $− 0.511+1.57i$ $0.511− 1.57i$ $0.511+1.57i$

Arjun asked in Calculus Feb 19, 2017

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gateme-2014-set4
calculus
definite-integrals
complex-variables

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piyag476 asked in Calculus Feb 19, 2017

GATE ME 2013 | Question: 26
The following surface integral is to be evaluated over a sphere for the given steady velocity vector field $F$ = $xi$ + $yj$+ $zk$ defined with respect to a Cartesian coordinate system having $i$, $j$ and $k$ ... is the outward unit normal vector to the sphere. The value of the surface integral is $\pi$ $2\pi$ $3\pi/4$ $4\pi$

piyag476 asked in Calculus Feb 19, 2017

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gateme-2013
calculus
integrals
area-under-curve

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go_editor asked in Calculus Mar 1, 2021

GATE Mechanical 2021 Set 2 | Question: 26
The value of $\int_{0}^{^{\pi }/_{2}}\int_{0}^{\cos\theta }r\sin\theta \:dr\:d\theta$ is $0$ $\frac{1}{6}$ $\frac{4}{3}$ $\pi$

go_editor asked in Calculus Mar 1, 2021

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gateme-2021-set2
calculus
definite-integrals
double-interals

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Arjun asked in Calculus Feb 9, 2019

GATE2019 ME-1: 2
A parabola $x=y^2$ with $0 \leq x \leq 1$ is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by $360^{\circ}$ around x-axis is $\dfrac{\pi}{4} \\$ $\dfrac{\pi}{2} \\$ ${\pi} \\$ $2 \pi$

Arjun asked in Calculus Feb 9, 2019

by Arjun

27.4k points

gateme-2019-set1
calculus
area-under-curve

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GATE Mechanical 2014 Set 1 | Question: 26

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