Triple Integral ( IS THIS RIGHT?)

[een]

Triple Integral ( IS THIS RIGHT??)

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\geq0, y\geq0. Compute\int\int\intxdxdydz

Homework Equations

The Attempt at a Solution

I did \int0-1\int0-1\int(x^2+y^2)-2 xdzdydx

\int0-1\int0-12x-x^3-xy^2 dydx
\int0-1[2xy-x^3y-xy^3/3]0-1 dx
\int0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
Can someone tell me if this is the correct answer

 

[lanedance]

pretty hard to read, below might help your formatting...
\int_{0}^1dx\int_{0}^1dy

 

[kandelabr]

you got x and y's upper limits wrong.
you are calculating the volume between these two surfaces (z(x,y) = x² + y², z < 2), see attachment.

 

[een]

okay.. this is what i got...
the lower limit of x is 0 the upper limit is sqrt(2)
same for y
the lower limit of z is x^2+y^2 and the upper limit is 2...
is that right?

 

[kandelabr]

yes, my result is 4/3.

 

[een]

i am getting sqrt(2)/3... what am i doing wrong?... i used the limits that you told me and I am pretty sure that i integrated correctly... can you show me step by step

 

[Mark44]

I have switched the order of integration, leaving the integrand unchanged. I think these are the correct limits of integration.
\int_{x = 0}^{\sqrt{2}} \int_{y = 0}^{\sqrt{2 - x^2}} \int_{z = x^2 + y^2}^{2} x~ dz dy dx

The projection of this region onto the x-y plane is the disk x^2 + y^2 <= 2, a disk of radius sqrt(2) centered at the origin. Geometrically, we have a stack of blocks dx by dy that extend from the paraboloid up to the plane z = 2. Then we let the stack sweep out along the y-axis from y = 0 to y = sqrt(2 - x^2) (i.e., out to the boundary of the disk), then finally let this wall of blocks sweep from x = 0 to x = sqrt(2).