Question

What is the volume of the solid generated by revolving the region bounded by the graphs y= sqrt(x), y = 0, and x = 4, about the line x = 6?

Answer 1

I've tried doing this using the equation\[V=\int\limits \pi f(y)^2dy\] and \[V=\int\limits \pi f(x)^2dx\] after adjusting the graph to be y= sqrt(x+6) [by saying that the y-axis is the line x= -6] but I can't get the right answer.

Answer 2

Sorry, the second eqn should be\[V = \int\limits 2 \pi x f(x) dx\]

Answer 3

again i forgot how to do this stuff, but let's put little help if possible
you boundaries are x=4 to x=6 yes?

Answer 4

did you try graphing this?

Answer 5

it would be clear and easier to go with

Answer 6

The boundaries would be from -6 to -2 on the adjusted graph (where the original x=6 is the new y-axis)

Answer 7

(x=-6 to x=-2)

Answer 8

how did you get those boundaries?

Answer 9

Well, it has to be x=-6 because a sqrt graph like that doesn't exist past that value. And in the original graph, the other boundary was stated to be at x=4, which would be x=-2 if we treated x=6 as the y-axis.