250+ TOP MCQs on Triple Integral and Answers

Differential and Integral Calculus Multiple Choice Questions 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
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}.)

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
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
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[zbig]_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) (logsqrt{2} – frac{5}{16})
d) (log⁡sqrt{4} – frac{6}{32})
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} = logsqrt{2} – frac{5}{16}.)

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
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}^zBig[xy + frac{y^2}{2} + zyBig]_{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}^1big[z(2x^2)+(2z^2 x)big]_{x=0}^z ,dz = int_{z=-1}^1 2z^3+2z^3 dz=big[z^4big]_{-1}^1=0.)

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