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Test: Double And Triple Integrals - 2 - Mathematics MCQ


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20 Questions MCQ Test - Test: Double And Triple Integrals - 2

Test: Double And Triple Integrals - 2 for Mathematics 2024 is part of Mathematics preparation. The Test: Double And Triple Integrals - 2 questions and answers have been prepared according to the Mathematics exam syllabus.The Test: Double And Triple Integrals - 2 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Double And Triple Integrals - 2 below.
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Test: Double And Triple Integrals - 2 - Question 1

The value of  dxdy is

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 1


[change the order]

Test: Double And Triple Integrals - 2 - Question 2

The volume of an object expressed in spherical coordinates is given by 

Q. The value of the integral is

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 2

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Test: Double And Triple Integrals - 2 - Question 3

Consider the shaded triangular region P shown in the figure. What is xy dxdy?

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 3

The intercept form of equation is


where, a is x intercept and b is y intercept.
So Given, equation is 
implies x+2y=2
implies x=2(1 - y)
Limit of x is form 0 to 2(1 - y ) and limit of y is from 0 to 1.
Therefore, 

= 1/6
 

Test: Double And Triple Integrals - 2 - Question 4

Changing the order of integration in the double integral leads to, then the value of q is

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 4

Given, I =  
Equation of the line is x = Ay Starting val ue of x is zero and final value of x is 8 but we want the result of first integration in terms of y, so that we can integrate the result with respect to y to get the final result. So, limits are from 0 to 4y.
And integration limit for y is from 0 to 2.

So, 

Hence, r=0, s=2, p=0, q=4y

Test: Double And Triple Integrals - 2 - Question 5

dxdy over the positive quadrant of the circle x2 + y2 = 1 is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 5

Let 
Changing to polar coordinates let x=r cosθ, y=r sinθ


Therefore,
= 1/14

Test: Double And Triple Integrals - 2 - Question 6

The volume of the cylinder x2 + y2 = a2 bounded below by z = 0 and bounded above by z = h is given by 

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 6

The equation of the cylinder is
x2 + y2 = a2
The equation of surface CDE is z = h
So, the required volume is


Let x = a sin θ
implies dx = a cosθ dθ
So, volume V

Test: Double And Triple Integrals - 2 - Question 7

dxdydz is equal to

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 7

Test: Double And Triple Integrals - 2 - Question 8

dydxdz is equal to

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 8


Test: Double And Triple Integrals - 2 - Question 9

If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then is equal to

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 9

Let I = 
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by


 

Test: Double And Triple Integrals - 2 - Question 10

If A is the region bounded by the parabolas y2 = 4x and x2 = 4y, then dxdy is equal to 

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 10

Let I = 
Solving the given equations
y2 =4x and x2 = 4y, we get x = 0, x = 4 The region of integration A is given by


 

Test: Double And Triple Integrals - 2 - Question 11

The area of the lemniscate r2 = a2 cos2θ is given by 

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 11

The curve is r2 = a2 cos 2θ
The required area

Test: Double And Triple Integrals - 2 - Question 12

The area bounded by the curves y2 = 9x, x —y + 2 = 0 is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 12

The equations of the given curves are

y2 = 9x
x - y + 2 = 0
The curves (/') and (ii) intersect at
A(1, 3) and B(4, 6)
So, the required area

= 1/2
 

Test: Double And Triple Integrals - 2 - Question 13

The area of the cardioid r = a (1 + cosθ) is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 13

The equation of the cardioid is

r = a (1 + cosθ)
Required area

Test: Double And Triple Integrals - 2 - Question 14

dydx is equal to

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 14

Test: Double And Triple Integrals - 2 - Question 15

The area bounded by the curves y = 2√x, y = - x, x = 1 and x = 4 is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 15

The given equations of the curves are

So, the required area is

Test: Double And Triple Integrals - 2 - Question 16

rdθdr is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 16


The equations of given curves are
y(x2 + 2) = 3x
and 4y = x2
The curve (i) and (ii) intersect at A(2,1). 

Test: Double And Triple Integrals - 2 - Question 17

The area o f the region bounded by the curve y(x2 + 2 ) = 3x and 4y = x2 is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 17

The required area


Test: Double And Triple Integrals - 2 - Question 18

The value of 

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 18

The value of dydx


 

Test: Double And Triple Integrals - 2 - Question 19

The area bounded by the curve r = θ cosθ and the lines θ = 0 and θ = π/2  is given by

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 19

The equation of the given curve is 
r =θcosθ -------(i)
So, the required area

Test: Double And Triple Integrals - 2 - Question 20

The solution of the differential equation dy/dx = x log x is

Detailed Solution for Test: Double And Triple Integrals - 2 - Question 20

Given, dy = x log x dx
⇒ y = x2log x/2 − ∫ x/2 dx [integrating]
⇒ y = (x2/2) log x − x2/4 + c

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