Ordinary Differential Equations Multiple Choice Questions on “Applications of Triple Integral”.
1. Evaluate ∫∫∫ 12y-8x dV in the region behind y=10-2z and bounded by z=2x, z=5 and x=0.
a) 1
b) (frac{35}{63})
c) (frac{3125}{16})
d) (frac{3125}{6})
Answer: d
Explanation: We know
From the boundary conditions,
0 < y < 10-2z
0 < x < (frac{z}{2})
0 < z < 5
Applying these limits on the Triple Integral as follows
(intintint_{0}^{10-2z} 12y-8x dy dx dz)
(=intint_{0}^{frac{pi}{2}}6 (10-2z)^2-8x(10-2z) dx dz)
(=∫_{0}^{5} 14z^3-130z^2+300z dz)
(=frac{3125}{6})
Thus the answer is (=frac{3125}{6}).
2. Assume a planet having a radius R and its density is expressed as = (frac{R+r}{2r}D’).
a) (frac{5pi D’R^3}{2})
b) (frac{4pi D’R^3}{3})
c) (frac{5pi D’R^3}{3})
d) (frac{5pi D’R^3}{12})
Answer: c
Explanation: Consider the case of r=R
Then,
D=D’
Where D’ is the surface density of the planet
As D → ∞, r → 0
For finding the mass of the planet, we use the triple integration formula
M=∫∫∫ dV
Converting into spherical co-ordinates, we get
M=(intintint D’r^2 sintheta frac{R+r}{2r} dr dtheta dtheta)
Applying the limits
0 to π
0 to 2π
0 to R
Solving the Triple Integral we get,
M=(frac{5pi D’R^3}{3})
Thus, the mass of the planet is (frac{5pi D’R^3}{3}).
Global Education & Learning Series – Ordinary Differential Equations.
To practice all areas of Ordinary Differential Equations,