Test: Double And Triple Integrals - 2 - Mathematics MCQ

[change the order]

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
 

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

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


Therefore,
= 1/14

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

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


 

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


 

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

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
 

The equation of the cardioid is

r = a (1 + cosθ)
Required area

The given equations of the curves are

So, the required area is

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

The required area

The value of dydx


 

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

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