Question:
Let
S = {(x, y, z) ∈ \(\R^3\) : x2 + y2 + z2 < 1}.
Then, the value of
\(\frac{1}{\pi}\iiint_s\left((x-2y+z)^2+(2x-y-z)+(x-y+2z)^2\right)dxdydz\)
equals ________ (rounded off to two decimal places).
Let
S = {(x, y, z) ∈ \(\R^3\) : x2 + y2 + z2 < 1}.
Then, the value of
\(\frac{1}{\pi}\iiint_s\left((x-2y+z)^2+(2x-y-z)+(x-y+2z)^2\right)dxdydz\)
equals ________ (rounded off to two decimal places).
S = {(x, y, z) ∈ \(\R^3\) : x2 + y2 + z2 < 1}.
Then, the value of
\(\frac{1}{\pi}\iiint_s\left((x-2y+z)^2+(2x-y-z)+(x-y+2z)^2\right)dxdydz\)
equals ________ (rounded off to two decimal places).
Updated On: Jan 25, 2025
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Correct Answer: 4.7
Solution and Explanation
The problem involves evaluating a triple integral over the unit sphere \( S \). By simplifying the integrand and using symmetry properties of the sphere, the result of the integral is approximately 4.70 when rounded to two decimal places.
Thus, the correct answer is 4.70.
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