Find volume of curve using disk method

1. about the line x = 6. (Give your answer correct to the nearest whole number.)
y=6 - x, y=0, y=2, x=0

2. about the x-axis. (Give your answer correct to 1 decimal place.)
y= x√(4-x²) , y=0

Please help me with these problems

all you got to do is draw the picture.
so first graph y=6-x y=0 and y=2 and x=0. So you are in the first quadrant only. Now you are being roatated around x=6. So this line is parallel to the y-axis so your bounds have to be from y=0 to y=2.
now we have to solve y=6-x in terms of y. x=6-y.
so your radius^2 is (6-(6-y))^2
so pi*integral[y^2du] from 0 to 2

1/3*y^3 from 0 to 2.
plug in 2 you get 8pi/3. plug in 0 you get 0.
Answer: 8pi/3.
2. your radius is x*sqrt(4-x^2). so
radius^2=x^2(4-x^2).
now you need to know where you are integratin from.
set x*sqrt(4-x^2)=0
x=0, x=2, x=-2. So you can use symmetry. integrate from 0 to 2 and multiply the integral by 2.
2pi*integral[4x^2-x^4dx] from 0 to 2
2pi[(4/3)x^3-(1/5)x^5] from 0 to 2
plug in 2
2pi[32/3 - 32/5]
2pi[160/15 - 96/15]
2pi[64/15]=128pi/15.
if you plug in 0 you get 0.
Answre: 128pi/15.