Find the volume V of the region bounded by z = 240-60y, z = 0, y = 0, and y = 4 -x2
To find the volume V of the region bounded by the given surfaces, we need to integrate the function z = 240 - 60y over the suitable region in the y-z plane.
Step 1: Determine the limits of integration.
The region is bounded by z = 0, y = 0, and y = 4 - x^2. We want to find the values of y and z that define the boundaries of this region.
Since z = 0 is a plane, it intersects the surface z = 240 - 60y at z = 0.
Setting z = 240 - 60y to zero, we find y = 4.
So, the limits of integration for y are from y = 0 to y = 4.
Next, we need to express y in terms of x. From the equation y = 4 - x^2, we have x^2 = 4 - y, which gives x = ±√(4 - y).
We can see that y ranges from 0 to 4, so x will range from -2 to 2.
Step 2: Set up the integral to find the volume.
The volume element (dV) in this case is represented as dV = dz * dy. We need to integrate this volume element over the region to find the total volume V.
V = ∫∫R (240 - 60y) * dy * dz
Step 3: Evaluate the integral.
Integrating dz first, we have:
V = ∫[0 to 4] ∫[0 to z = 0] (240 - 60y) dy dz
Simplifying the inner integral, we have:
V = ∫[0 to 4] [240y - 30y^2] | [0 to z = 0] dz
V = ∫[0 to 4] (240z - 30z^2) dz
Integrating with respect to z, we get:
V = [120z^2 - 10z^3] | [0 to 4]
V = (120(4)^2 - 10(4)^3) - (120(0)^2 - 10(0)^3)
V = (120(16) - 10(64)) - (0 - 0)
V = 1920 - 640
V = 1280 cubic units
Therefore, the volume of the region bounded by the given surfaces is 1280 cubic units.