Question

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by y=x^3 and y=9x about the x-axis.

Answer 1

first solve \[x^3=9x\]

Answer 2

x(x^2-9)=0

Answer 3

yes

Answer 4

so x=?

Answer 5

x=0 at 0 and 3?

Answer 6

0, -3 ,3 ....but we can ignore the -3 since it is not in the first quadrant.

Answer 7

so our integral is going to go from x=0 to x=3

Answer 8

between 0 and 3 , which function is on top?

Answer 9

x^3 should be and then subtract 9x?

Answer 10

no

Answer 11

pick any number between 0 and 3...like 1

and plug into each function
1^3=1
9*1=9
9>1

Answer 12

so 9x is on top of x^3 between 0 and 3

Answer 13

now we can use the formula


\[V=\int_{a}^{b}\pi f(x)^2dx\]

Answer 14

so from 0,3 pi(9x)^2- pi(x^3)^2= VOLUME?

Answer 15

\[\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx\]

yes

Answer 16

so (pi(9x^3/3)-pi(x^7/7))dx

Answer 17

no

Answer 18

\[\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx\]

\[=\pi\int\limits_{0}^{3}(81x^2-x^6)dx\]

Answer 19

the volume is\[2\times \int\limits_{0}^{3}[\pi(9x)^2-\pi(x^3)^2]dx\]

Answer 20

we are working in the first quadrant

Answer 21

so i won't have to take the antiderivative?

Answer 22

you will....you forgot to square your 9

Answer 23

\[=\pi\left[81\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{3}\]

Answer 24

\[=\pi\left[81\frac{3^3}{3}-\frac{3^7}{7}\right]\]

Answer 25

and 2×∫ 0 3 [π(9x) 2 −π(x 3 ) 2 ]dx =5832pi/7

Answer 26

you don't need the times 2.

Answer 27

416.5714pi

Answer 28

\[=\frac{2916\pi}{7}\]

Answer 29

yes

Answer 30

thank you so much

Answer 31

no problem

Answer 32

?the graph is adove x-axis and below x-axis,it is symetric,so i have to times 2

Answer 33

no you don't

Answer 34

all we need it the area of a circle

\[\pi r^2\]

if f(x) is the height of the function then

we have

\[\pi f(x)^2\]

we can then multiply by the thickness of our disks and add them all up

V=\[\int\limits_{a}^{b}\pi (f(x))^2dx\]


no 2 times in this formula.