The plane z + y = 1 {\displaystyle \displaystyle z+y=1} intesects z = 0 {\displaystyle \displaystyle z=0} at the line y = 1 , z = 0 {\displaystyle \displaystyle y=1,z=0} . Since the region is is the first octant, we know that x , y , z ≥ 0 {\displaystyle \displaystyle x,y,z\geq 0} .
Using this information and the sketch in the hint, we find the region E {\displaystyle \displaystyle E} as E = { ( x , y , z ) ∈ R 3 | 0 ≤ z ≤ 1 − y , x 2 ≤ y ≤ 1 , 0 ≤ x ≤ 1 } {\displaystyle \displaystyle E=\{(x,y,z)\in \mathbb {R} ^{3}|\ 0\leq z\leq 1-y,\ x^{2}\leq y\leq 1,\ 0\leq x\leq 1\}}
Hence the integral is
∭ E x d V = ∫ 0 1 ∫ x 2 1 ∫ 0 1 − y x d z d y d x = ∫ 0 1 ∫ x 2 1 x ( 1 − y ) d y d x = ∫ 0 1 x [ y − y 2 2 ] x 2 1 d x = ∫ 0 1 x 2 − x 3 + x 5 2 d x = [ x 2 4 − x 4 4 + x 6 12 ] 0 1 = 1 12 {\displaystyle \displaystyle {\begin{aligned}\iiint _{E}x\;{\text{d}}V&=\int _{0}^{1}\int _{x^{2}}^{1}\int _{0}^{1-y}x\;{\text{d}}z{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}\int _{x^{2}}^{1}x(1-y)\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}x\left[y-{\frac {y^{2}}{2}}\right]_{x^{2}}^{1}\;{\text{d}}x\\&=\int _{0}^{1}{\frac {x}{2}}-x^{3}+{\frac {x^{5}}{2}}\;{\text{d}}x\\=&\left[{\frac {x^{2}}{4}}-{\frac {x^{4}}{4}}+{\frac {x^{6}}{12}}\right]_{0}^{1}={\frac {1}{12}}\end{aligned}}}
intesects 
at the line 
. Since the region is is the first octant, we know that 
.
as
![{\displaystyle \displaystyle {\begin{aligned}\iiint _{E}x\;{\text{d}}V&=\int _{0}^{1}\int _{x^{2}}^{1}\int _{0}^{1-y}x\;{\text{d}}z{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}\int _{x^{2}}^{1}x(1-y)\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}x\left[y-{\frac {y^{2}}{2}}\right]_{x^{2}}^{1}\;{\text{d}}x\\&=\int _{0}^{1}{\frac {x}{2}}-x^{3}+{\frac {x^{5}}{2}}\;{\text{d}}x\\=&\left[{\frac {x^{2}}{4}}-{\frac {x^{4}}{4}}+{\frac {x^{6}}{12}}\right]_{0}^{1}={\frac {1}{12}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/837f2610e2135e7698a0c6b894af32123885bd78)
