Question

The region bounded by the curve y=sqrt(x) and the lines x=0 and y=6 rotated about the line x=-2

Answer 1

What have you tried?
What method do you prefer?

Answer 2

Washer method

Answer 3

Okay, then set it up: \(\pi \int r^{2}\;dh\)

Go!

Answer 4

Isnt there an inner and outer radius?

Answer 5

There is, but the inner radius is zero (0).

Answer 6

Oh, sorry, I read x = 0 and x = 6. Rethink and find the inner radius.

x = 0 and y = 6!

Answer 7

?

Answer 8

You will need the intersection of y = 6 and y = sqrt(x).

Answer 9

ya (6,36)

Answer 10

Okay, the Outer Radius is a constant 8. What is the Inner Radius?

Answer 11

should be 2

Answer 12

It is 2 ONLY for x = 0. How about for the rest of the Domain, (0,36]?

Answer 13

i have no clue

Answer 14

The you cannot solve this problem.

Give it a go. What is the inner radius for x = 1? \(\sqrt{1} = 1\), so the Inner Radius there would be 1-(-2) = 1+2 = 3

How about x = 4?

Answer 15

4

Answer 16

How did you get that?

Answer 17

does 8-y^2 have something to do with it

Answer 18

You didn't answer by question and no.

Answer 19

i literally just plugged in 4 for x

Answer 20

Oh, so you used the actual function to determine the radius. That's it. Your inner radius is \(\sqrt{x}\).

Okay, now the limits.

Answer 21

Oh, so you used the actual function to determine the radius. That's it.

Your inner radius is \(\sqrt{x}+2\).

Okay, now the limits.

Answer 22

one should be 36

Answer 23

oops, 6

Answer 24

the other maybe -2

Answer 25

The radius is parallel to the y-axis. Your limits need to be on the x-axis, perpendicular to your radius.

Answer 26

so 0 and 36

Answer 27

Done. What do you get?

Answer 28

i got 936 but i know thats not right

Answer 29

You missed the \(\pi\).

Answer 30

Alternatively, with Shells

\(2\pi\int\limits_{0}^{6}(y+2)\cdot y^{2}\;dy\) Same thing.

Answer 31

u sure that is right?

Answer 32

the 936pi

Answer 33

That's why I like to do it both ways. If I don't get the same answer both ways, I keep working at it.

Yes!

Answer 34

the outer radius was 8 and inner sqrt(x)+2 correct?

Answer 35

Indeed.

Answer 36

\(\pi\int\limits_{0}^{36}8^{2} - (\sqrt{x}+2)^{2}\;dx\)

Answer 37

my online thing says its wrong

Answer 38

oh well thanks brother

Answer 39

It's possible we don't have the right problem statement. This is the solution to the problem presented.

Answer 40

the region bounded by the curve y=sqrt(x) and the lines x=0 and y=6. Then it asks me to find the volume of the solid when the region is rotated about the line x= -2

Answer 41

I believe you. That is the problem we solved.

I cannot speak for the online acceptance. There is an error in this process, somewhere. It is not in the solution to the problem presented.

Does it want a purely numerical response? The numbers in this problem are rather large. There may be some rounding problem.

Try 2940.5290987556 and 2940.5307237601

Answer 42

ill ask my math teacher about it tuesday, i LOVE high school ap calc

Answer 43

Go get 'em.

The more I see online attempts at math education, without real humans, the less I am impressed. Maybe one day we'll start to get better at it.

Answer 44

true that man, true that