This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses 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) \(\frac{3125}{6}\)
View Answer
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
\(\int\int\int_{0}^{10-2z} 12y-8x dy dx dz\)
\(=\int\int_{0}^{\frac{\pi}{2}}6 (10-2z)^2-8x(10-2z) dx dz\)
\(=∫_{0}^{5} 14z^3-130z^2+300z dz\)
\(=\frac{3125}{6}\)
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) \(\frac{5\pi D’R^3}{2}\)
b) \(\frac{4\pi D’R^3}{3}\)
c) \(\frac{5\pi D’R^3}{3}\)
d) \(\frac{5\pi D’R^3}{12}\)
View Answer
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=\(\int\int\int D’r^2 sin\theta \frac{R+r}{2r} dr d\theta d\theta\)
Applying the limits
0 to π
0 to 2π
0 to R
Solving the Triple Integral we get,
M=\(\frac{5\pi D’R^3}{3}\)
Thus, the mass of the planet is \(\frac{5\pi D’R^3}{3}\).
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
To practice all areas of Ordinary Differential Equations, here is complete set of 1000+ Multiple Choice Questions and Answers.
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