Differential and Integral Calculus Questions and Answers – Triple Integral

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This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Triple Integral”.

1. The value of \(\int_0^1 \int_0^x \int_0^{x+y} \,xyz \,dz\, dy\, dx\,\) is given by ____
a) 17/144
b) 16/72
c) 17/72
d) 15/144
View Answer

Answer: a
Explanation: \(\int_{x=0}^1 \int_{y=0}^x \int_{z=0}^{x+y}\,xyz \,dz\, dy\, dx\,\)
\(\int_{x=0}^1 \int_{y=0}^x xy[\frac{z^2}{2}]_0^{x+y} \,dy \,dx = \int_{x=0}^1 \int_{y=0}^x xy \frac{(x+y)^2}{2} \,dy \,dx\)
\(=\frac{1}{2} \int_{x=0}^1 \int_{y=0}^x xy(x^2 + y^2 + 2xy) \,dy \,dx\)
\(=\frac{1}{2} \int_{x=0}^1 \int_{y=0}^x(x^3 y + xy^3 + 2x^2 y^2) \,dy \,dx\)
\(=\frac{1}{2} \int_{x=0}^1 \Big[x^3 \frac{y^2}{2} + x \frac{y^4}{4} + \frac{2x^2 y^3}{3}\Big]_{y=0}^x dx = \frac{1}{2} \int_{x=0}^1(\frac{x^5}{2} + \frac{x^5}{4} + \frac{2x^5}{3}) \,dx\)
\(=\frac{1}{2} \Big[\frac{x^6}{12} + \frac{x^6}{24} + \frac{x^6}{9}\Big]_{x=0}^1=(\frac{1}{12} + \frac{1}{24} + \frac{1}{9})\frac{1}{2} = \frac{17}{144}.\)
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2. The integral value of \(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z} \,dz \,dy \,dx\) is given by _____
a) \(=\frac{1}{3}(e^{4a}+6e^{2a}+8e^{a}+3)\)
b) \(=\frac{1}{3}(e^{4a}-6e^{2a}+4e^{a}+3)\)
c) \(=\frac{1}{8}(e^{4a}-6e^{2a}+8e^{a}-3)\)
d) 0
View Answer

Answer: c
Explanation: \(\int_0^a \int_0^x \int_0^{x+y} e^{x+y+z} \,dz \,dy \,dx = \int_0^a \int_0^x \int_0^{x+y} e^{x+y} e^{z} \,dz \,dy \,dx\)
\(\int_0^a \int_0^x e^{x+y} [e^{z}]_0^{x+y} \,dy \,dx = \int_0^a \int_0^x e^{x+y} (e^{x+y}-1) \,dy \,dx\)
\(\int_0^a \int_0^x(e^{2x+2y}-e^{x+y}) \,dy \,dx = \int_0^a {e^{2x} \big[\frac{e^{2y}}{2}\big]_0^x – e^x [e^y]_0^x} \,dx\)
\(\int_0^a(\frac{e^{4x}}{2} – \frac{3}{2} e^{2x} + e^x)dx = \big[\frac{e^{4x}}{8} – \frac{3}{4} e^{2x} + e^x \big]_0^a \)
\(= (\frac{e^{4a}}{8} – \frac{3}{4} e^{2a} + e^a)-(\frac{1}{8}-\frac{3}{4}+1)\)
\(=\frac{1}{8}(e^{4a}-6e^{2a}+8e^{a}-3)\).

3. The integral value of \(\int_0^{\frac{π}{2}} \int_0^{a sinθ} \int_0^r r \,dr \,dθ \,dz \) is _____
a) 0.5
b) 0.25
c) 1
d) 0
View Answer

Answer: d
Explanation: \(\int_{θ=0}^{\frac{π}{2}} \int_{r=0}^{a sinθ} \int_{z=0}^r r \,dr \,dθ \,dz = \int_{θ=0}^{\frac{π}{2}} \int_{r=0}^{a sinθ} r \big[z\big]_0^r \,dr \,dθ \)
\(= \int_{θ=0}^{\frac{π}{2}} \int_{r=0}^{a sinθ} r^2 \,dr \,dθ\)
\(\int_{θ=0}^{\frac{π}{2}}\big[\frac{r^3}{3}\big]_0^{sin⁡θ} \,dθ = \int_0^{\frac{π}{2}} \frac{sin^3 θ}{3} \,dθ = \int_0^{\frac{π}{2}} \frac{3 sin⁡θ-sin⁡3θ}{12} \,dθ = \Big[\frac{-3 cos⁡θ + 3 cos⁡3θ}{12}\Big]_0^{\frac{π}{2}}=0\).

4. The integral value of \(\int_0^1 \int_0^{1-x} \int_0^{1-x-y} \frac{dz dy dx}{(1+x+y+z)^3} \) is given by_____
a) \(log⁡\sqrt{2} – \frac{7}{16}\)
b) \(log⁡\sqrt{4} + \frac{5}{32}\)
c) \(log\sqrt{2} – \frac{5}{16}\)
d) \(log⁡\sqrt{4} – \frac{6}{32}\)
View Answer

Answer: c
Explanation: \(\int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{1-x-y} \frac{dz dy dx}{(1+x+y+z)^3} = \int_0^1 \int_0^{1-x}\Big[\frac{-1}{(2(1+x+y+z)^2}\Big]_{z=0}^{1-x-y} \,dy \,dx\)
\(\int_0^1 \int_0^{1-x}[\frac{-1}{8} + \frac{1}{(2(1+x+y)^2)}] \,dy \,dx = \int_0^1 \Big[\frac{-y}{8} – \frac{1}{2(1+x+y)}\Big]_{y=0}^{1-x} \,dx\)
\(\int_{x=0}^1 \Big[\frac{-(1-x)}{8} – \frac{1}{4} + \frac{1}{2(x+1)}\Big]dx = \int_{x=0}^1 \Big[\frac{-3}{8} + \frac{x}{8} + \frac{1}{2(x+1)}\Big]dx\)
\(\Big[\frac{-3x}{8} + \frac{x^2}{16} + \frac{1}{2} log(x+1)\Big]_{x=0}^1 = \frac{3}{8} + \frac{1}{16} + \frac{log⁡2}{2} = log\sqrt{2} – \frac{5}{16}.\)
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5. The integral of \(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z} (x+y+z)\,dy \,dx \,dz\) is given by _______
a) 0
b) 1
c) 0.25
d) 4
View Answer

Answer: a
Explanation: \(=\int_{z=-1}^1 \int_{x=0}^z \int_{y=x-z}^{x+z}(x+y+z)dy \,dx \,dz = \int_{z=-1}^1 \int_{x=0}^z\Big[xy + \frac{y^2}{2} + zy\Big]_{y=x-z}^{x+z} \,dx \,dz\)
\( =\int_{z=-1}^1 \int_{x=0}^z \Big\{x((x+z)-(x-z))+\frac{1}{2} [(x+z)^2-(x-z)^2] \\
+z((x+z)-(x- z))\Big\}dx \,dz\)
\( =\int_{z=-1}^1 \int_{x=0}^z(2xz+2xz+2z^2)dx \,dz\)
\( = \int_{-1}^1\big[z(2x^2)+(2z^2 x)\big]_{x=0}^z \,dz = \int_{z=-1}^1 2z^3+2z^3 dz=\big[z^4\big]_{-1}^1=0.\)

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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