SOLUTION: What is the volume of solid in xyz-space bounded by surfaces y = x^2, y = 2 - x^2, z = 0 and z = y + 3?

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Question 1172398: What is the volume of solid in xyz-space bounded by surfaces y = x^2, y = 2 - x^2, z = 0 and z = y + 3?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let's find the volume of the solid.
**1. Determine the Region in the xy-Plane**
* We need to find the intersection of the parabolas y = x^2 and y = 2 - x^2.
* Set them equal: x^2 = 2 - x^2
* 2x^2 = 2
* x^2 = 1
* x = ±1
* When x = ±1, y = 1.
* The region in the xy-plane is bounded by these parabolas, with x ranging from -1 to 1.
**2. Set up the Triple Integral**
* The volume is given by the triple integral:
* V = ∫∫∫ dV
* The limits of integration are:
* z: 0 to y + 3
* y: x^2 to 2 - x^2
* x: -1 to 1
* The integral becomes:
* V = ∫(from -1 to 1) ∫(from x^2 to 2 - x^2) ∫(from 0 to y + 3) dz dy dx
**3. Evaluate the Integral**
* First, integrate with respect to z:
* ∫(from 0 to y + 3) dz = [z](from 0 to y + 3) = y + 3
* Now, integrate with respect to y:
* ∫(from x^2 to 2 - x^2) (y + 3) dy = [y^2/2 + 3y](from x^2 to 2 - x^2)
* = [(2 - x^2)^2/2 + 3(2 - x^2)] - [(x^2)^2/2 + 3(x^2)]
* = [4 - 4x^2 + x^4]/2 + 6 - 3x^2 - x^4/2 - 3x^2
* = 2 - 2x^2 + x^4/2 + 6 - 3x^2 - x^4/2 - 3x^2
* = 8 - 8x^2
* Finally, integrate with respect to x:
* ∫(from -1 to 1) (8 - 8x^2) dx = [8x - 8x^3/3](from -1 to 1)
* = [8(1) - 8(1)^3/3] - [8(-1) - 8(-1)^3/3]
* = [8 - 8/3] - [-8 + 8/3]
* = 8 - 8/3 + 8 - 8/3
* = 16 - 16/3
* = (48 - 16)/3
* = 32/3
**Answer**
The volume of the solid is 32/3 cubic units.
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
What is the volume of solid in xyz-space bounded by surfaces y = x^2, y = 2 - x^2, z = 0 and z = y + 3?
~~~~~~~~~~~~~~~~~~~~~~


                    I will solve this problem mentally. 


In (x,y)-plane, the area concluded between y = x^2, x-axis, and 0 <= x <= 1 is 1/3.

It is elementary calculation from Calculus.


From it, we deduce, that the area concluded between y = x^2, y = 1  and 0 <= x <= 1 is 2/3.


Hence, the area concluded between y = x^2 and y = 2-x^2 is 4 times 2/3, or 8/3 square units.


Now, our 3D solid consists of two parts.


One part is a right cylinder 0 <= z <= 3 over its base, which the area concluded 
between y = x^2 and y = 2-x^2.


The volume of this cylinder is   = 8 cubic units.



The other part is half of the cylinder  3 <= z <= 5  with the same base.

The height of this imaginary cylinder is 2 units (z from 3 to 5), so, its volume is  = 16/3.


The plane z = z + 3 cuts this cylinder in two parts of equal volumes - it is clear from the symmetry.


So, the whole volume of the 3D body under the interest is   =  =  = 32/3 cubic units.

Solved mentally.

            My interior voice tells me that this mental solution
                    is what is expected in this problem.



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