Answer:
13.5
Step-by-step explanation:
We are given that two curves and a line
[tex]y=x^2,x=y^2[/tex] about y=-6
[tex]x=\sqrt{y}[/tex]
Height=[tex]\sqrt{y}-y^2[/tex]
Radius=y-(-6)=y+6
Now, we find intersecting points of two curves
[tex]y=(y^2)^2[/tex]
[tex]y=y^4[/tex]
[tex]y^4-y=0[/tex]
[tex]y(y^3-1)=0[/tex]
[tex]y=0[/tex]
[tex]y^3-1=0[/tex]
[tex]y^3=1\implies y=1[/tex]
Now, volume generated by rotating the region by using cylindrical shell method is given by
[tex]V=\int_{a}^{b}2\pi(height)(radius)dy[/tex]
Using the formula
[tex]V=\int_{0}^{1}2\pi(\sqrt{y}-y^2)(y+6)dy[/tex]
[tex]V=\int_{0}^{1}2\pi(y^{\frac{3}{2}}+6\sqrt{y}-y^3-6y^2)dy[/tex]
[tex]V=2\pi[\frac{2}{5}y^{\frac{5}{2}}+4y^{\frac{3}{2}}-\frac{1}{4}y^4-2y^3]^{1}_{0}[/tex]
[tex]V=2\pi(\frac{2}{5}+4-\frac{1}{4}-2)[/tex]
[tex]V=13.5 [/tex]
