# Ordinary Differential Equations Questions and Answers – Applications of Triple Integral

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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) $$\frac{3125}{6}$$

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) $$\frac{5\pi D’R^3}{12}$$

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. 