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Triple Integral ( IS THIS RIGHT??)
Let R be the solid enclosed by the planes x=0, y=0, z=2 and the surface x^2+y^2, where
x[tex]\geq[/tex]0, y[tex]\geq[/tex]0. Compute[tex]\int\int\int[/tex]xdxdydz
I did [tex]\int[/tex]0-1[tex]\int[/tex]0-1[tex]\int[/tex](x^2+y^2)-2 xdzdydx
[tex]\int[/tex]0-1[tex]\int[/tex]0-12x-x^3-xy^2 dydx
[tex]\int[/tex]0-1[2xy-x^3y-xy^3/3]0-1 dx
[tex]\int[/tex]0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
Can someone tell me if this is the correct answer
Homework Statement
Let R be the solid enclosed by the planes x=0, y=0, z=2 and the surface x^2+y^2, where
x[tex]\geq[/tex]0, y[tex]\geq[/tex]0. Compute[tex]\int\int\int[/tex]xdxdydz
Homework Equations
The Attempt at a Solution
I did [tex]\int[/tex]0-1[tex]\int[/tex]0-1[tex]\int[/tex](x^2+y^2)-2 xdzdydx
[tex]\int[/tex]0-1[tex]\int[/tex]0-12x-x^3-xy^2 dydx
[tex]\int[/tex]0-1[2xy-x^3y-xy^3/3]0-1 dx
[tex]\int[/tex]0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
Can someone tell me if this is the correct answer