Question 1172398
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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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                    I will solve this problem mentally.



<pre>
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  {{{(8/3)*3}}} = 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 {{{(8/3)*2}}} = 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  {{{8 + 16/6}}} = {{{8 + 8/3}}} = {{{(24+8)/3}}} = 32/3 cubic units.
</pre>

Solved mentally.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;My interior voice tells me that this mental solution

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;is what is expected in this problem.