Differential and Integral Calculus Questions and Answers – Change of Variables In a Triple Integral

«
»

This set of 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 ______
differential-integral-calculus-questions-answers-mcqs-q1
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
View Answer

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
\(f(ρ,∅,z)*J\left(\frac{x,y,z}{ρ,∅,z}\right)\) with limits in functions of x,y,z to functions of ρ,∅,z respectively
\(J\left(\frac{x,y,z}{ρ,∅,z}\right)= \begin{vmatrix}
\frac{∂x}{∂p} & \frac{∂x}{∂∅} &\frac{∂x}{∂z}\\
\frac{∂y}{∂p} &\frac{∂y}{∂∅} &\frac{∂y}{∂z}\\
\frac{∂z}{∂p} &\frac{∂z}{∂∅} &\frac{∂z}{∂z}\\
\end{vmatrix}
= \begin{vmatrix}
cos⁡∅ &-p sin⁡∅ &0\\
sin⁡∅ &p cos⁡∅ &0\\
0 &0 &1\\
\end{vmatrix} = cos ∅(ρ cos ∅) + ρ sin ∅ (sin ∅)\)
= ρ, thus ∭R f(x,y,z)dx dy dz = ∭R* f(ρ,∅,z) ρ dρ d∅ dz where R* is the new region.
advertisement

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 ______
differential-integral-calculus-questions-answers-mcqs-q2
a) ∭R* f(r,θ,∅) sin⁡θ dr dθ d∅
b) ∭R* f(r,θ,∅) r2 dr dθ d∅
c) ∭R* f(r,θ,∅) r2 cos⁡θ dr dθ d∅
d) ∭R* f(r,θ,∅) r2 sin⁡θ dr dθ d∅
View Answer

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 \(f(ρ,∅,z)*J\left(\frac{x,y,z}{ρ,∅,z}\right)\) with limits in functions of x,y,z to functions of r,θ,∅ respectively
\(J\left(\frac{x,y,z}{ρ,∅,z}\right) =\begin{vmatrix}
\frac{∂x}{∂r} &\frac{∂x}{∂θ} &\frac{∂x}{∂∅}\\
\frac{∂y}{∂r} &\frac{∂y}{∂θ}& \frac{∂y}{∂∅}\\
\frac{∂z}{∂r} &\frac{∂z}{∂θ} &\frac{∂z}{∂∅}\\
\end{vmatrix} = \begin{vmatrix}
sin θ cos ∅ &r cos θ cos ∅ &-r sin θ sin ∅\\
sin θ sin ∅ &r cos θ sin ∅ &r sin θ cos ∅\\
cos θ& -r sin θ &0\\
\end{vmatrix}\)
= 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 sin⁡θ
thus ∭R f(x,y,z)dx dy dz = ∭R* f(r,θ,∅)r2 sin⁡θ 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
View Answer

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
View Answer

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
\(J\left(\frac{p,q,r}{x,y,z}\right) = \begin{vmatrix}
\frac{∂p}{∂x} & \frac{∂p}{∂y} &\frac{∂p}{∂z}\\
\frac{∂q}{∂x} &\frac{∂q}{∂y} &\frac{∂q}{∂z}\\
\frac{∂r}{∂x} &\frac{∂r}{∂y} &\frac{∂r}{∂z}\\
\end{vmatrix} = \begin{vmatrix}
1&1&1\\
0&1&1\\
0&0&1\\
\end{vmatrix} = 1(1) – 1(0) + 1(0) = 1\)
but we need \(J\left(\frac{x,y,z}{p,q,r}\right) \,w.k.t\, J\left(\frac{x,y,z}{p,q,r}\right) J\left(\frac{p,q,r}{x,y,z}\right) = 1 \,thus\, J\left(\frac{x,y,z}{p,q,r}\right)=1\)
now the volume is given by \(\int_ 0^1 \int_ 0^2\int_ 0^3 \,dp \,dq \,dr = \int_ 0^1 \int_ 0^2 3\, dq \,dr = \int_ 0^1 6dr = 6.\)
advertisement

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) \(\frac{8π(e+1)}{3}\)
View Answer

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 sin⁡θ 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
\(\displaystyle\int_ 0^{2π} \int_ 0^π \int_ 0^1 e^{{r}^{{2}^{1.5}}} r^2 \, sin⁡θ \,dr \,dθ \,d∅ = \int_ 0^{2π} d∅ \int_ 0^π sin⁡θ \,dθ \int_ 0^1 r^2 e^{{r}^{3}} \,dr\)
\(2π*\Big[-cos⁡θ\Big]_0^π * \frac{1}{3} \Big[e^{{r}^{3}}\Big]_{r^3=0}^{r^3=1}=\frac{4π(e-1)}{3}.\)

Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.

To practice MCQs on all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn