Question

Find the volume of the solid obtained by rotating the region bounded by y=x^3 and x=y^3 in the first quadrant about the x-axis.

Answer 1

Can you set up the definite integral that leads to this answer?

Answer 2

no im totally lost

Answer 3

it is multiple choice

Answer 4

There must be limits of sorts...

Answer 5

from 0 to 1, perhaps?

Answer 6

that is the only thing the question says

Answer 7

I see. Well, I'll assume it's from 0 to 1, then, because that is where they intersect, ok? :)

Answer 8

ok

Answer 9

Well, you realise that from 0 to 1, x = y^3 is higher up than y = x^3, right?

Answer 10

yes

Answer 11

the answer can be 16/35pi, 16/7pi, 18/35, 7/2, or 16pi

Answer 12

In other words, we write both of them as functions of x... like so
\[f(x) = \sqrt[3]{x}\]
and
\[g(x) = x^{3}\]

And from 0 to 1, f(x) is greater than g(x)

Answer 13

ok

Answer 14

so what do we do next?

Answer 15

If f(x) is always greater than or equal to g(x) in the interval, then the volume of the solid of rotation is given by this formula (I can do my best to derive it if you so wish)

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

Answer 16

Where (a,b) is your interval :)
Can you evaluate the integral?

Answer 17

yes please

Answer 18

yes please what?

Answer 19

oh that was it.. what do we do from there?:

Answer 20

Well, what's the interval, it's [0,1], right? and we have our values for f and g so all that's left is to plug them in :)

Answer 21

how do we plug a and b into x?

Answer 22

your values for a and b are 0 and 1, respectively, as we have that interval :)

Answer 23

Are you evaluating it?

Answer 24

\[\int \left(\pi \sqrt[3]{x}^2-\pi
\left(x^3\right)^2\right) \, dx=\pi
\left(\frac{3
x^{5/3}}{5}-\frac{x^7}{7}\right)+K

\]

Answer 25

Evaluate the above from 0 to 1 to get
\[
\int_0^1 \left(\pi \sqrt[3]{x}^2-\pi
\left(x^3\right)^2\right) \, dx=\frac{16
\pi }{35}
\]