How do you find the volume of the solid bounded by Z = 1 – y^2, x + y = 1, and the three coordinate plane?

Question

Scarlett Allison · Accepted Answer

#5/12# plugging the limits into #V = int_V \ dV = int_V dx \ dy \ dz#, we have #V = int_(x = 0)^1 int_(y=0)^(1-x) int_(z = 0)^(1 - y^2) dz \ dy \ dx# # = int_(x = 0)^1 int_(y=0)^(1-x) 1 - y^2 \ dy \ dx# # = int_(x = 0)^1 [ y - y^3/3 ]_(0)^(1-x) \ dx# # = int_(x = 0)^1 1-x -(1-x)^3/3 \ dx# # = [ x-x^2/2 + (1-x)^4/12 ]_0^1 # # = [ 1-1 /2 ] - [1/12] = 5/12#

Christopher Allers · Answer

undefined To find the volume of the solid bounded by the surfaces $ z = 1 - y^2 $, $ x + y = 1 $, and the three coordinate planes, you can use triple integration.

The bounds for the volume integral will be determined by the intersection of the given surfaces. The region of intersection lies within the limits of $ x $, $ y $, and $ z $.

First, find the limits of integration for $ z $. Since the upper surface is $ z = 1 - y^2 $, and the lower surface is the xy-plane (z = 0), the limits for $ z $ will be from 0 to $ 1 - y^2 $.

Next, determine the limits for $ y $. These are bounded by the curve $ x + y = 1 $. Solving for $ y $, you get $ y = 1 - x $. The limits for $ y $ will thus be from 0 to $ 1 - x $.

Finally, the limits for $ x $ are determined by the region of intersection between the surfaces. Since it's a simple triangle in the xy-plane, the limits for $ x $ will be from 0 to 1.

So, the volume integral will be:

$$ V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-y^2} dz \, dy \, dx $$

You can then compute this triple integral to find the volume of the solid bounded by the given surfaces.