- #1
carl123
- 56
- 0
Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid
x = 5y2+ 5z2 and the plane x = 5.
My work so far:
Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y2+5z2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Also, when x=0, y2 + z2=0 -----> r=0 and x=5 -----> y2+z2=1 -----> r=1, so my bounds for r are 0 ≤r ≤ 1.
My triple integral is then:
∫∫∫E 5x dV = ∫(from 0 to 2π) ∫(from 0 to 1) ∫(from 5r2 to 5) 5xr dx dr dθ
I keep getting the wrong answer after evaluating it
x = 5y2+ 5z2 and the plane x = 5.
My work so far:
Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y2+5z2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Also, when x=0, y2 + z2=0 -----> r=0 and x=5 -----> y2+z2=1 -----> r=1, so my bounds for r are 0 ≤r ≤ 1.
My triple integral is then:
∫∫∫E 5x dV = ∫(from 0 to 2π) ∫(from 0 to 1) ∫(from 5r2 to 5) 5xr dx dr dθ
I keep getting the wrong answer after evaluating it