Volume under surface

Calculus Level 2

A 3D surface is described by

z = e − ( x 2 + y 2 ) z = e^{-(x^2 + y^2) }

Find the volume of the region between this surface and the x y x y -plane.

e − 1 e^{-1} π \pi π 2 \dfrac{\pi}{\sqrt{2}} e \sqrt{e}

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Tom Engelsman
Nov 2, 2018

Easy one with cylindrical coordinates. Let r 2 = x 2 + y 2 , r ∈ [ 0 , ∞ ) , θ ∈ [ 0 , 2 π ] . r^2 = x^2 + y^2, r \in [0, \infty), \theta \in [0, 2\pi]. The volume just computes according to:

V = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ = π . V = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r dr d\theta = \boxed{\pi}.

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