Differential and Integral Calculus MCQs focuses on “Change of Variables In a Triple Integral”.
1. For the below-mentioned figure, conversion from cartesian coordinate ∭R f(x,y,z)dx dy dz to cylindrical polar with coordinates p(ρ,∅,z) is given by ______
a) ∭R* f(ρ,∅,z) ρ dρ d∅ dz
b) ∭R f(ρ,∅,z) dρ d∅ dz
c) ∭R*f(ρ,∅,z) ρ∅ dρ d∅ dz
d) ∭R f(ρ,∅,z) ρ2 dρ d∅ dz
Answer: a
Explanation: From the figure we can write x=ρ cos ∅, y=ρ sin ∅, z=z
now we know that during change of variables f(x,y,z) is replaced by
(f(ρ,∅,z)*Jleft(frac{x,y,z}{ρ,∅,z}right)) with limits in functions of x,y,z to functions of ρ,∅,z respectively
(Jleft(frac{x,y,z}{ρ,∅,z}right)= begin{vmatrix}
frac{∂x}{∂p} & frac{∂x}{∂∅} &frac{∂x}{∂z}\
frac{∂y}{∂p} &frac{∂y}{∂∅} &frac{∂y}{∂z}\
frac{∂z}{∂p} &frac{∂z}{∂∅} &frac{∂z}{∂z}\
end{vmatrix}
= begin{vmatrix}
cos∅ &-p sin∅ &0\
sin∅ &p cos∅ &0\
0 &0 &1\
end{vmatrix} = cos ∅(ρ cos ∅) + ρ sin ∅ (sin ∅))
= ρ, thus ∭R f(x,y,z)dx dy dz = ∭R* f(ρ,∅,z) ρ dρ d∅ dz where R* is the new region.
2. For the below mentione figure ,conversion from cartesian coordinate ∭R f(x,y,z)dx dy dz to spherical polar with coordinates p(r,θ,∅) is given by ______
a) ∭R* f(r,θ,∅) sinθ dr dθ d∅
b) ∭R* f(r,θ,∅) r2 dr dθ d∅
c) ∭R* f(r,θ,∅) r2 cosθ dr dθ d∅
d) ∭R* f(r,θ,∅) r2 sinθ dr dθ d∅
Answer: d
Explanation: From the figure we can write x = r sin θ cos ∅, y = r sin θ sin ∅, z = r cos θ
now we know that during a change of variables f(x,y,z) is replaced by (f(ρ,∅,z)*Jleft(frac{x,y,z}{ρ,∅,z}right)) with limits in functions of x,y,z to functions of r,θ,∅ respectively
(Jleft(frac{x,y,z}{ρ,∅,z}right) =begin{vmatrix}
frac{∂x}{∂r} &frac{∂x}{∂θ} &frac{∂x}{∂∅}\
frac{∂y}{∂r} &frac{∂y}{∂θ}& frac{∂y}{∂∅}\
frac{∂z}{∂r} &frac{∂z}{∂θ} &frac{∂z}{∂∅}\
end{vmatrix} = begin{vmatrix}
sin θ cos ∅ &r cos θ cos ∅ &-r sin θ sin ∅\
sin θ sin ∅ &r cos θ sin ∅ &r sin θ cos ∅\
cos θ& -r sin θ &0\
end{vmatrix})
= sin θ cos ∅(r2 sin2 θ cos∅) + r cos θ cos ∅(r sin θ cos ∅ cos θ) – r sin θ sin ∅
= (-r sin2 θ sin∅-r cos2 θ sin∅)……on solving we get r2 sinθ
thus ∭R f(x,y,z)dx dy dz = ∭R* f(r,θ,∅)r2 sinθ dr dθ d∅ where R* is the new region.
3. If ∭R xyz dx dy dz is solved using cylindrical coordinate where R is the region bounded by the planes x=0, y=0, z=0, z=1 & x2+y2=1 then what is the value of that integral?
a) 1/24
b) 1/16
c) 1/4
d) 1/2
Answer: b
Explanation: x2+y2=1→ρ varies from 0 to 1 substituting x=ρ cos ∅, y=ρ sin ∅, z=z
z varies from 0 to1, x=0, y=0→∅ varies from 0 to π/2
thus the given integral is changed to cylindrical polar given by
(int_0^{frac{π}{2}} int_ 0^1 int_0^1 cos∅sin∅ ρ^3 z ,dz ,dρ ,d∅ = int_0^{frac{π}{2}} int_ 0^1 cos∅sin∅ ρ^3 Big[frac{z^2}{2}Big]_0^1 ,dρ ,d∅)
(int_0^{frac{π}{2}} cos∅sin∅ Big[frac{ρ^3}{8}Big]_0^1 ,d∅ = int_0^{frac{π}{2}} cos∅sin∅ frac{1}{8} ,d∅ )
put sin ∅=t, dt=cos ∅
t varies from 0 to 1 (int_ 0^1 frac{1}{8} t ,dt = Big[frac{t^2}{16}Big]_0^1 = frac{1}{16}.)
4. The volume of the region R defined by inequalities 0≤z≤1, 0≤y+z≤2,0≤x+y+z≤3 is given by ______
a) 4
b) 6
c) 8
d) 1
Answer: b
Explanation: It is observed from equations that the region is made of parallelepiped thus volume of parallelepiped is given by triple integral over the given region.
i.e by using substitutions as x+y+z=p, y+z=q, z=r the new region becomes R* where p varies from 0 to 3, q varies from 0 to 2 & r varies from 0 to 1 jacobian of this transformation is given by
(Jleft(frac{p,q,r}{x,y,z}right) = begin{vmatrix}
frac{∂p}{∂x} & frac{∂p}{∂y} &frac{∂p}{∂z}\
frac{∂q}{∂x} &frac{∂q}{∂y} &frac{∂q}{∂z}\
frac{∂r}{∂x} &frac{∂r}{∂y} &frac{∂r}{∂z}\
end{vmatrix} = begin{vmatrix}
1&1&1\
0&1&1\
0&0&1\
end{vmatrix} = 1(1) – 1(0) + 1(0) = 1)
but we need (Jleft(frac{x,y,z}{p,q,r}right) ,w.k.t, Jleft(frac{x,y,z}{p,q,r}right) Jleft(frac{p,q,r}{x,y,z}right) = 1 ,thus, Jleft(frac{x,y,z}{p,q,r}right)=1)
now the volume is given by (int_ 0^1 int_ 0^2int_ 0^3 ,dp ,dq ,dr = int_ 0^1 int_ 0^2 3, dq ,dr = int_ 0^1 6dr = 6.)
5. What is the value of integral (∭_Re^{{(x^2+y^2+z^2)}^{frac{3}{2}}} ,dx ,dy ,dz ) where R is the region given by x2+y2+z2≤1?
a) (frac{4π(e-1)}{3})
b) (frac{4π(e^3-1)}{3})
c) (frac{4π(e^2+1)}{3})
d) (frac{8π(e+1)}{3})
Answer: a
Explanation: It can be noticed that R is the region bounded by sphere from the equation x2+y2+z2≤1 thus we are using spherical coordinate to solve this problem
i.e clearly radius r varies from 0 to 1, θ varies from 0 to π & ∅ varies from 0 to 2π
thus the given integral changes to (displaystyle∭_{R^*} e^{{r}^{{2}^{1.5}}}) r2 sinθ dr dθ d∅
(e^{{r}^{{2}^{1.5}}}) is obtained by substituting x = r sin θ cos ∅, y = r sin θ sin ∅, z=r cos θ & hence solving the same, now substituting R* we get
(displaystyleint_ 0^{2π} int_ 0^π int_ 0^1 e^{{r}^{{2}^{1.5}}} r^2 , sinθ ,dr ,dθ ,d∅ = int_ 0^{2π} d∅ int_ 0^π sinθ ,dθ int_ 0^1 r^2 e^{{r}^{3}} ,dr)
(2π*Big[-cosθBig]_0^π * frac{1}{3} Big[e^{{r}^{3}}Big]_{r^3=0}^{r^3=1}=frac{4π(e-1)}{3}.)
Global Education & Learning Series – Differential and Integral Calculus.
To practice MCQs on all areas of Differential and Integral Calculus,