T think.ms.tink New member Joined Apr 3, 2009 Messages 18 Apr 30, 2009 #1 Evaluate the triple integral{E} (x + 4 y) dV where E is bounded by the parabolic cylinder y = 2 x^2 and the planes z = 5 x, y = 2 x, and z = 0.
Evaluate the triple integral{E} (x + 4 y) dV where E is bounded by the parabolic cylinder y = 2 x^2 and the planes z = 5 x, y = 2 x, and z = 0.
D DrMike Full Member Joined Mar 31, 2009 Messages 251 May 1, 2009 #2 think.ms.tink said: Evaluate the triple integral{E} (x + 4 y) dV where E is bounded by the parabolic cylinder y = 2 x^2 and the planes z = 5 x, y = 2 x, and z = 0. Click to expand... Please explain what you've tried so far, and where you are stuck.
think.ms.tink said: Evaluate the triple integral{E} (x + 4 y) dV where E is bounded by the parabolic cylinder y = 2 x^2 and the planes z = 5 x, y = 2 x, and z = 0. Click to expand... Please explain what you've tried so far, and where you are stuck.
T think.ms.tink New member Joined Apr 3, 2009 Messages 18 May 1, 2009 #3 setting the limits of integration of 0 to 1, 2x^2 to 2x and 0to 5x of x+4y dz dy dx and for my answer i got (50/4-10)... which was incorrect
setting the limits of integration of 0 to 1, 2x^2 to 2x and 0to 5x of x+4y dz dy dx and for my answer i got (50/4-10)... which was incorrect
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 May 2, 2009 #4 It appears you have the correct integral. You must have evaluated incorrectly. I get 23/6. \(\displaystyle \int_{0}^{1}\int_{2x^{2}}^{2x}\int_{0}^{5x}(x+4y)dzdydx=\frac{23}{6}\)
It appears you have the correct integral. You must have evaluated incorrectly. I get 23/6. \(\displaystyle \int_{0}^{1}\int_{2x^{2}}^{2x}\int_{0}^{5x}(x+4y)dzdydx=\frac{23}{6}\)