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

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

\(\int\int\int_{0}^{10-2z} 12y-8x dy dx dz\)

\(=\int\int_{0}^{\frac{\pi}{2}}6 (10-2z)^2-8x(10-2z) dx dz\)

\(=∫_{0}^{5} 14z^3-130z^2+300z dz\)

\(=\frac{3125}{6}\)

Thus the answer is \(=\frac{3125}{6}\).

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 \(=\frac{3125}{6}\).

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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

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=\(\int\int\int D’r^2 sin\theta \frac{R+r}{2r} dr d\theta d\theta\)

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 \(\frac{5\pi D’R^3}{3}\).

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.**

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