Differential and Integral Calculus Questions and Answers for Experienced people focuses on “Change of Variables In a Double Integral”.
1. Evaluation of (intint_R f(x,y) ,dx ,dy ) in cartesian coordinate can be done using change of variables principle, among the choices given below which is correct explanation of change of variables principle? (Given let x=g(u,v) & y=h(u,v))
a) (intint_S f(g(u,v),h(u,v)) ,du ,dv)
b) (intint_S f(g(u,v),h(u,v)) frac{d(x,y)}{d(u,v)} ,du ,dv)
c) (intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv)
d) (intint_S f(g(u,v),h(u,v)) frac{∂(u,v)}{∂(x,y)} ,du ,dv)
Answer: c
Explanation: (intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv)
where ( frac{∂(x,y)}{∂(u,v)} = begin{vmatrix}
frac{∂x}{∂u} frac{∂x}{∂v}\
frac{∂y}{∂u} frac{∂y}{∂v}\
end{vmatrix} = J(frac{x,y}{u,v}) = frac{∂x}{∂u} frac{∂y}{∂v} – frac{∂x}{∂v} frac{∂y}{∂u} …..)(‘J’ is Jacobian).
2. The value of ∬R (x-y)2 dx dy where R is the parallelogram with vertices (0,0), (1,1),(2,0), (1,-1) when solved using change of variables is given by____
a) 16/3
b) 8/3
c) 4/3
d) 0
Answer: b
Explanation: W.K.T from change of variables principle
(intint_R f(x,y),dx ,dy = intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv …..(1))
From the above diagram in the region R the equations are given by
x-y=0, x-y=2, x+y=0, x+y=2 from this we can observe that change of
variables is u=x-y, v=x+y solving we get (x=frac{u+v}{2}, y=frac{v-u}{2})
(frac{∂(x,y)}{∂(u,v)} = begin{vmatrix}
frac{∂x}{∂u}&frac{∂x}{∂v}\
frac{∂y}{∂u}&frac{∂y}{∂v}\
end{vmatrix} = begin{vmatrix}
0.5 &0.5\
-0.5&0.5\
end{vmatrix} = 0.5)
The region S in the (u,v) is the square 0
(=int_0^2 Big[frac{u^3}{6}Big]_0^2 ,dv = int_0^2 frac{4}{3} dv = frac{8}{3}.)
3. If double integral in Cartesian coordinate is given by ∬R f(x,y) dx dy then the value of same integral in polar form is _____
a) ∬P f(r cos θ, rsin θ)dr dθ
b) ∬P f(r cosθ, r sinθ)rdr dθ
c) ∬P f(r cosθ, r sinθ) r2 dr dθ
d) ∬P f(r sinθ, r cosθ)dr dθ
Answer: b
Explanation: ∬R f(x,y)dx dy when converting this into polar form we take x = r cos θ
y=r sin θ as change of variables from
(intint_R f(x,y) ,dx ,dy = intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv ) where u=r & v=θ
thus (frac{∂(x,y)}{∂(r,θ)} =begin{vmatrix}
frac{∂x}{∂r} &frac{∂x}{∂θ}\
frac{∂y}{∂r} &frac{∂y}{∂θ}\
end{vmatrix} = begin{vmatrix}
cosθ &-r sinθ\
sinθ & rcosθ\
end{vmatrix} = r(cos^2 θ + sin^2 θ) = r)
substituting we get ∬P f(r cos θ,r sin θ)rdr dθ.
4. The value of ∬R sin(x2 + y2) dx dy where R is the region bounded by circle centered at origin with radius r=2 is _____
a) πcos 4
b) π(1-cos 4)
c) π
d) π(1-sin 4)
Answer: b
Explanation: Using Polar variable transformation x = r cos θ & y=r sin θ, r varies from 0 to 2 & θ varies from 0 to 2π because radius of circle i.e r=2 & centered at origin
(intint_P f(r cos θ,r sin θ )rdr ,dθ = intint_R sin (x^2+y^2) ,dx ,dy = int_0^2π int_0^2 r sin r^2 ,dr ,dθ)
Using substitution t=r2 integral changes to (int_0^{2π} int_0^4 0.5 ,sint ,dt ,dθ)
(int_0^{2π} 0.5big[-costbig]_0^4 dθ = int_0^{2π} 0.5(1-cos4) dθ = π(1-cos 4)).
5. Using change of variables principle in double integral we can reduce cartesian integral to simpler form.
a) True
b) False
Answer: a
Explanation: The above statement is not necessarily true always but change of variables cartesian to polar form & changing into suitable form reduces the complexity in the evaluation of double integral however converse is also true for example ∬R x2 y3 dx dy is the given integral while solving there is no need of converting it into polar form since ∬P r6 cos2 θ sin3 θ dr dθ is tedious to solve when R only consists of constants but ∬R x2 y3 dx dy can be solved easily by ordinary method of integration.
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