Question

Would someone be willing to check my answer for the following problem?

x = y^3, x = y^2; determine the area between the curves from y=0 to 1.
For a final answer I got |-1/12| units^2.

Any and all help is greatly appreciated! :)

Answer 1

I set it up as an integral following (f1(y) - f2(y))dy, with x=y^3 being f1(y). I'm pretty sure this is how it should be set up, but my answer being negative and a kind of weird fraction threw me off.

Answer 2

http://www.wolframalpha.com/input/?i=area+between+x+%3D+y%5E3+and+x+%3D+y%5E2+from+y%3D0+to+y%3D1

Answer 3

Thank you!!!!

Answer 4

If you got a negative, you had the functions the wrong way around.

Answer 5

Okay... :/ would x=y^2 be the "top" function in this case then?

Answer 6

Even though, when graphed, it is on the bottom within the interval we are evaluating [0, 1] ??

Answer 7

x = y^3, x = y^2; determine the area between the curves from y=0 to 1
that is exactly the same as
y = x^3, y = x^2; determine the area between the curves from x=0 to 1

swapping the x's and y's changes nothing here, really.

Answer 8

So it is like the graph gets flipped on its side, in a way, when we evaluate it in terms of x? And I should have set up the functions equal to y first? Is that the safe route to go?

Answer 9

there's no need to change them to y =, i was just pointing out that if x^2 is on top, why wouldn't y^2 be on top?

Answer 10

Okay... I think I see what your saying. On the graph from WolframAlpha though, it looks like the blue line (x=y^2) is physically on top during the part needing evaluated... Is that an incorrect understanding?

Answer 11

You're evaluating with respect to y, the vertical axis.

Answer 12

y^2 is on top, from 0 to 1

Answer 13

So, I look at it in regard to what is farther from the axis which the limits apply to/ are on? Since x=y^2 would be in top if the whole graph were rotated 90 degrees CCW, we just look at it that way, with the function axis we are evaluating in respect to being the one we determine "top" and "bottom" from?

Answer 14

I think I follow the rational of how to determine that now, for problems now just in terms of x.... I think so.

Answer 15

Don't need to rotate 90 degrees... x is increasingly positive as you go to the right. Further right = x is bigger.

Answer 16

Same reasoning as for y - as you go higher up, y increases.

Answer 17

Not from 0 to 1 though... :/

Answer 18

...what? I'm talking in general. Up = increasing y. Right = increasing x.

Answer 19

Okay, I just thought that the one on top within the limits had to go first in the integral. I'll talk to my Calc teacher tomorrow about this problem as well. Thank you for your help! :) Sorry to keep you so long:/

Answer 20

"I just thought that the one on top within the limits had to go first in the integral"

that's true. y^2 is on top. The integral is y^2 - y^3, respect to y.

Answer 21

I'm not quite sure what's confusing you... x = y^2 is greater than, or above, x = y^3 between 0 and 1