Find the volume in the first octant bounded by y ^2 = 4x, 2x + y = 4, z = y, and y = 0. Ans. 5/3?
I repeatedly tried this but I cannot arrive at this answer. Please help, I'm at my wits end here.
- anonymousLv 71 month ago
I did my best to explain my thought process. The integration itself is too lengthy to type out in this format, so I attached a screenshot of an online calculator for that.
The volume is in the first octant, so x ≥ 0, y ≥ 0, and z ≥ 0.
We've got a boundary equation z = y, so we can use 0 ≤ z ≤ y as one of three inequalities to describe the volume. (Integrating from the xy-plane, z = 0, to the plane z = y.)
Now we need to describe the region in the xy-plane to integrate over. I graphed the equations y² = 4x and 2x + y = 4 to see the region. (picture attached).
From that picture, integrating with respect to x first and then with respect to y is the way to go. We need the previous two boundary equations as functions of y instead of x.
y² = 4x ⟶ x = y²/4
2x + y = 4 ⟶ x = (4 - y) / 2
The region in the xy-plane can be described as
0 < y < 2
y²/4 < x < (4 - y) / 2
In the attached picture, the region in the xy-plane that we're interested in is the wedge shape near the origin.
volume integral is
∫ [y from 0 to 2] ∫ [x from y²/4 to (4 - y) / 2] ∫ [z from 0 to y] dz dx dy