Question

volume problem

Answer 1

take the integral from 1-4 of your R which is the equation they give you but squared and times it by pi

Answer 2

This volume problem is an integral. Do you understand what a cross section looks like

Answer 3

integral from 1 to 4

Answer 4

yeah ik the bounds are 1 to 4 but idk what to put in the integral equation

Answer 5

its just the equation they give you but the whole equation is squared so factor it out and then integrate but dont forget the PI

Answer 6

4x^2- x^2 what

Answer 7

\[\pi \int\limits_{1}^{4} (4x-x^2)^2 dx \]

Answer 8

OOH ok is there a reason why its squared tho

Answer 9

its just the way the formula is for these revolution problems its R^2 or R^2-r^2 depending on if there is a whole in the revolution or not

Answer 10

hole*

Answer 11

oh ok thank you so much ! :)

Answer 12

That's what the graph is and if you rotate this about the x-axis each cross section is a circle. What is the area of each circle? pi*r^2, and what is r? radius is given by the height from the x-axis to the curve above...so it changes as you go from x=1 to x=4. Thus your integral you are adding up pi*r^2 and r=given equation, hence the integral above.

Answer 13

no problem im in Calculus right now so glad to share my knowledge helps me prepare :)

Answer 14

:))) hahah it truly did

Answer 15

are u in AP?

Answer 16

yeah im in ap lol