# 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.

### 1 Answer

- anonymousLv 74 months 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.

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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.

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volume integral is

∫ [y from 0 to 2] ∫ [x from y²/4 to (4 - y) / 2] ∫ [z from 0 to y] dz dx dy