Find the volume V of the solid in the �rst octant bounded by y = 0, z = 0, y = 3,
z = x, and z + x = 4.
The best part about this problem is that I solved it using basic geometry but am expected to do it using a triple integral that will take up a whole page.
I'm honestly lost on how im supposed to set up the integral I've tried setting it up various ways and i never really even come close to the volume 12, which it is.
Looks like a triangular prism, with base a 3x4 rectangle in the xy plane, and top edge a line parallel to the y-axis at x=2, z=2.
I also get v=12 using simple geometry.
Integrating in two pieces, one under z=x, and the other under z=4-x, we get
∫[0,3]∫[0,2]x dxdy + ∫[0,3]∫[2,4](4-x)dx dy
= ∫[0,3](1/2 x^2)[0,2]dy + ∫[0,3](4x-x^2/2)[2,4]dy
= ∫[0,3]2dy + ∫[0,3]2dy
= 6+6
= 12
To find the volume of the solid bounded by the given surfaces, we can indeed use a triple integral. Let's break down the problem step by step to set up the integral correctly.
1. Sketch the region: It's always helpful to visualize the region in order to understand how it is bounded. In this case, we are working in the first octant, so the region lies in the positive x, y, and z axes.
2. Identify the limits of integration: Since the region is bounded by four surfaces, we need to determine the limits for each of the three variables (x, y, and z) in the integral.
a. For x: We are given z + x = 4. Solving for x, we get x = 4 - z. The limits for x will be determined by this equation, which means x ranges from 0 to 4 - z.
b. For y: We are given the y = 0 and y = 3. Therefore, y ranges from 0 to 3.
c. For z: We are given z = 0 and z = x. This means z ranges from 0 to x.
3. Set up the triple integral: The volume V can be expressed as the triple integral of 1 with respect to x, y, and z over the given limits:
V = ∫∫∫ 1 dV
= ∫∫∫ dx dy dz
= ∫₀³ ∫₀⁴-z ∫₀ˣ dx dy dz
4. Evaluating the triple integral: We integrate with respect to x first, then y, and finally z.
V = ∫₀³ ∫₀⁴-z ∫₀ˣ dx dy dz
= ∫₀³ ∫₀⁴-z [x]₀ˣ dy dz (Integrating x from 0 to 4 - z)
= ∫₀³ ∫₀⁴-z (4 - z) dy dz
= ∫₀³ [(4 - z) * y]₀⁴-z dz (Integrating y from 0 to 4 - z)
= ∫₀³ (4 - z)(4 - z) dz
= ∫₀³ (16 - 8z + z^2) dz
= [16z - 4z^2 + (z^3)/3]₀³
= (16(3) - 4(3^2) + (3^3)/3) - (16(0) - 4(0^2) + (0^3)/3)
= 36 - 36/3
= 36 - 12
= 24
Therefore, the volume V of the solid bounded by the given surfaces in the first octant is 24 cubic units, not 12 as mentioned in your question.