Question

Suppose R is the region in the first quadrant bounded by the curves y=3/x y=(x^2)/9 the volume of the solid resulting when revolving R about the y−axis.

Answer 1

You wanna graph the curves first?

Answer 2

ok so i feel like i can confidently do it if it we're around the Y axis, but that x=1 is what is throwing me off

Answer 3

What do you mean \(x=1\)?

Answer 4

lol oh wow, i never typed that part in

Answer 5

Revised :Suppose R is the region in the first quadrant bounded by the curves y=3/x y=(x^2)/9 and x=1 . Find the volume of the solid resulting when revolving R about the y−axis.

Answer 6

Let's graph first.

Answer 7

I would let \[
\frac 3x=\frac{x^2}{9}
\]

Answer 8

Simplifying that seems to get \(x=3\), when they intersect.

Answer 9

yeah i have the same thing

Answer 10

So our \(x\) limits are between \(1\) from \(x=1\) and \(3\).

Answer 11

Yup, i'm on board so far

Answer 12

I suppose we should do shell method

Answer 13

\[
2\pi\int r(x)h(x)\;dx
\]In this case \(r(x)=x\) and \(h(x)=3/x-x^2/9\)

Answer 14

Does that make sense to you?

Answer 15

Yeah, but when i'm solving for the Y-axis doesn't the x=1 play a role in the equation?

Answer 16

Well

Answer 17

and , if you don't mind me asking, is there certain scenarios where shell method is preferred over say, the washer method