Calculus III triple integral Help!
Favorites|Homepage
Subscriptions | sitemap
HOME > > Calculus III triple integral Help!

Calculus III triple integral Help!

[From: ] [author: ] [Date: 12-04-03] [Hit: ]
1] ∫[0, x^2] ∫[0,= ∫[-1, 1] ∫[0,= ∫[-1,= (x^3)/3 - (x^5)/10 eval.......
Use a triple integral to find the volume of the region bounded by y = x^2, z = 0, and y + z = 1 in the order dz dy dx

-
Volume = ∫[-1, 1] ∫[0, x^2] ∫[0, 1 - y] dz dy dx

= ∫[-1, 1] ∫[0, x^2] (1 - y) dy dx

= ∫[-1, 1] x^2 - (x^4)/2 dx

= (x^3)/3 - (x^5)/10 eval. from -1 to 1

= 1/3 - 1/10 + 1/3 - 1/10 = 2/3 - 1/5 = 7/15
1
keywords: Help,III,integral,triple,Calculus,Calculus III triple integral Help!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .