16. Changing the order of the integration in the double integral
    

    leads to

    .What is q ?

1. 1. 4y
      

   
   
   

   2. 16 y²
      

   
   
   

   3. x
      

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   When I =

   f(x , y) dydx

   
   
   

   I =

   f(x , y) dxdy

   Correct Option: A

   When I =

   f(x , y) dydx

   
   
   

   I =

   f(x , y) dxdy

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17. By a change of variable x(u, y) = uv, y(u, v) = v/u is double integral, the integrand f(x, y) changes to f(uv, v/u) φ (u, v). Then, φ(u, v) is
    

1. 1. 2 v / u
      

   
   
   

   2. 2 uv
      

   
   
   

   3. v²
      

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: A

   

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18. The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of
    

1. 1. 1	m
      3

      
      

   
   
   

   2. 2	m
      3

      
      

   
   
   

   3. 2√2	m
      3

      
      

   
   
   

   4. 4	m
      3

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Given R = 1, radins of sphere.
   Let height of cone is H = h + R
   

   Volume , V =	1	π × (√R² - h²)2(R + h)
   	3

   
   for maximum value ,
   

   ⇒	dV	= 0
   	dh

   
   

   ⇒	d		1	(R² - h²)(R + h)
   	dh		3

   
   ⇒ -2h(R + h) + (R² - h²) = 0
   ⇒ (R + h)(R - 3h) = 0
   

   h = -R ,	R
   	3

   
   

   Height of the come = R +	R	=	4R
   	3		3

   
   

   =	4 × 1	m =	4	m
   	3		3

   
   

   Correct Option: D

   
   Given R = 1, radins of sphere.
   Let height of cone is H = h + R
   

   Volume , V =	1	π × (√R² - h²)2(R + h)
   	3

   
   for maximum value ,
   

   ⇒	dV	= 0
   	dh

   
   

   ⇒	d		1	(R² - h²)(R + h)
   	dh		3

   
   ⇒ -2h(R + h) + (R² - h²) = 0
   ⇒ (R + h)(R - 3h) = 0
   

   h = -R ,	R
   	3

   
   

   Height of the come = R +	R	=	4R
   	3		3

   
   

   =	4 × 1	m =	4	m
   	3		3

   
   

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

1. 1. 1

      6

      
      

   
   
   

   2. 2

      9

      
      

   
   
   

   3. 7

      16

      
      

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   I = ∬ xy .dxdy
   

   The limit of y is from 0 to

   and limit of x from 0 to 2.

   
   
   
   

   =		x		1 -	x		2	dx
   		2			2

   
   

   =	1		x(x² + 4 - 4x).dx
   	8

   
   

   =	1		(x³ + 4x - 4x²).dx
   	8

   
   

   Correct Option: A

   I = ∬ xy .dxdy
   

   The limit of y is from 0 to

   and limit of x from 0 to 2.

   
   
   
   

   =		x		1 -	x		2	dx
   		2			2

   
   

   =	1		x(x² + 4 - 4x).dx
   	8

   
   

   =	1		(x³ + 4x - 4x²).dx
   	8

   
   

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20. Which of the following intergrals is unbounded?
    

1. 1. tanx dx

      
      

   
   
   

   2. 	1	dx
      	x2 + 1

      
      

   
   
   

   3. xe-x dx

      
      

   
   
   

   4. 	1	dx
      	1 - x

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   To see whether the integrals are bounded or unbounded, we need to see that the integrand doesn’t r each i nfi ni t y i n t he i nt er val of integration. Let us write down the range of the integrands in the 4 options,
   a → (0, 1); b → (0, 1); c → (0, 1 / e ); d → (0, ∞)
   

   Thus ,(d) , i.e.

   dx is unbounded .

   Correct Option: D

   To see whether the integrals are bounded or unbounded, we need to see that the integrand doesn’t r each i nfi ni t y i n t he i nt er val of integration. Let us write down the range of the integrands in the 4 options,
   a → (0, 1); b → (0, 1); c → (0, 1 / e ); d → (0, ∞)
   

   Thus ,(d) , i.e.

   dx is unbounded .

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