Hi! I need help with the attached question in calc. Thank you:)

Answer:
3π square units.
Step-by-step explanation:
We can use the disk method.
Since we are revolving around AB, we have a vertical axis of revolution.
So, our representative rectangle will be horizontal.
R₁ is bounded by y = 9x.
So, x = y/9.
Our radius since our axis is AB will be 1 - x or 1 - y/9.
And we are integrating from y = 0 to y = 9.
By the disk method (for a vertical axis of revolution):
[tex]\displaystyle V=\pi \int_a^b [R(y)]^2\, dy[/tex]
So:
[tex]\displaystyle V=\pi\int_0^9\Big(1-\frac{y}{9}\Big)^2\, dy[/tex]
Simplify:
[tex]\displaystyle V=\pi\int_0^9(1-\frac{2y}{9}+\frac{y^2}{81})\, dy[/tex]
Integrate:
[tex]\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big][/tex]
Evaluate (I ignored the 0):
[tex]\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi[/tex]
The volume of the solid is 3π square units.
Note:
You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:
[tex]\displaystyle V=\frac{1}{3}\pi(1)^2(9)=3\pi[/tex]
We acquire the exact same answer.