Answer:
(a) and (b) see pictures attached
(c) V = 16/35
Step-by-step explanation:
(a) Sketch the base of S in the xy-plane.
See picture 1 attached
(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.
See picture 2 attached
(c) Compute the volume of S.
The volume is given by the triple integral
[tex]\displaystyle\iiint_{S}zdzdydx[/tex]
The cross-sections perpendicular to the x-axis are squares so
[tex]z=1-x^2[/tex]
The region S is given by the following inequalities
[tex]0\leq x\leq 1\\\\x^2\leq y\leq 1\\\\0\leq z\leq 1-x^2[/tex]
Therefore
[tex]\displaystyle\iiint_{S}zdzdydx=\displaystyle\int_{0}^{1} \displaystyle\int_{x^2}^{1} \displaystyle\int_{0}^{1-x^2} (1-x^2)dzdydx=\\\\\displaystyle\int_{0}^{1}(1-x^2)(1-x^2)(1-x^2)dx=\displaystyle\int_{0}^{1}(1-x^2)^3dx=\displaystyle\frac{16}{35}[/tex]
So the volume V of the solid S is
V=16/35
