Question

U

Answer 1

The area under the curve is parallel to the y-axis, so you will be using shell method.
A constant of 2pi will be in front of your integral, leaving rh to be determined.
the radius is , and your height is the area under the curve from 4 to 6.
So, let me take a stab at using the equation button:
\[\pi \int\limits_{4}^{6}x(x-2)dx\]
The /2 from x-2/2 and the 2 from 2pi cancel each other out.

Answer 2

surface area tends to involve sqrts

Answer 3

about the y axis\[\int2\pi x ~ds\]
and\[ds=\sqrt{(x')^2+(y')^2}~:(dx~or~dy)\]

Answer 4

y=x-2/2

2y+2 = x

ds = sqrt(2^2+1) dy

\[2\pi\int_{a}^{b}x~ds=2\pi\int_{a}^{b}(2y+2)~\sqrt{5}~dy\]

Answer 5

to find the y limits, use the x interval to find the needed y values