Confusing volume

Geometry Level 4

Find out the volume enclosed by the function f ( z ) = ∣ x + y ∣ f(z)=|x+y| for − 1 ≤ x ≤ 1 -1\leq x \leq1 and − 1 ≤ y ≤ 1 -1\leq y \leq 1 . Find the answer to 4 decimals after point.


The answer is 2.6667.

Alapan Das by Alapan Das , 20, India

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1 solution

Alapan Das
Jun 21, 2019

The volume V V = ∫ − 1 1 ∫ − 1 1 ∣ x + y ∣ d x d y \int_{-1}^1 \int_{-1}^{1} |x+y| dx dy = 2 ( ∫ 0 1 ∫ 0 1 ∣ x + y ∣ d x d y ) + 2 ( ∫ 0 1 ( ∫ 0 y ( y − x ) d x + ∫ y 1 ( x − y ) d x ) d y ) 2(\int_{0}^1 \int_{0}^1 |x+y| dx dy)+2(\int_{0}^1 (\int_{0}^y (y-x) dx +\int_{y}^1 (x-y) dx) dy) = 2 + 2 3 2+\frac{2}{3} = 2.6667 2.6667

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