Area between Curves: Find x=16 Solution

[Asphyxiated]

Homework Statement

[URL]http://images3a.snapfish.com/232323232%7Ffp733%3B%3B%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D33625%3B9674347nu0mrj[/URL]

Find the area s between the curves:

y^{2}=x

2y=x

x=16

Homework Equations

The Attempt at a Solution

So i am rather certain that I have the right area marked in the picture to find. It would be much simpler to find this area with respect to x instead of y so I will solve the equations for y and take the integral from the intersect to x=16.

y=\pm \sqrt{x}= \pm x^{1/2}

the positive side is all we need for this integral though.

y = \frac {x}{2}

and these two functions intersect and x=4. So the integral should be:

\int_{4}^{16} \frac {x}{2} - x^{1/2} dx

\frac {x^{2}}{4} - \frac {2x^{3/2}}{3}

which is the anti-derivative that needs to be evaluated at 16 and 4..

[64-\frac{128}{3}]-[4-\frac{16}{3}]

[\frac {64}{3}] - [-\frac{4}{3} ]

\frac {68}{3}

which is evidently not correct. I have tried this many different ways both with respect to x and y and with other areas as S. Like I stated before I am pretty sure I am trying to find the right area now but I am not having much luck here. Please help!

 

[Mark44]

Homework Statement

[URL]http://images3a.snapfish.com/232323232%7Ffp733%3B%3B%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D33625%3B9674347nu0mrj[/URL]

Find the area s between the curves:

y^{2}=x

2y=x

x=16

Homework Equations

The Attempt at a Solution

So i am rather certain that I have the right area marked in the picture to find. It would be much simpler to find this area with respect to x instead of y so I will solve the equations for y and take the integral from the intersect to x=16.

y=\pm \sqrt{x}= \pm x^{1/2}

the positive side is all we need for this integral though.

y = \frac {x}{2}

and these two functions intersect and x=4. So the integral should be:

\int_{4}^{16} \frac {x}{2} - x^{1/2} dx

\frac {x^{2}}{4} - \frac {2x^{3/2}}{3}

which is the anti-derivative that needs to be evaluated at 16 and 4..

[64-\frac{128}{3}]-[4-\frac{16}{3}]

[\frac {64}{3}] - [-\frac{4}{3} ]

\frac {68}{3}

which is evidently not correct. I have tried this many different ways both with respect to x and y and with other areas as S. Like I stated before I am pretty sure I am trying to find the right area now but I am not having much luck here. Please help!

The way I read this problem is that there are two regions - the first interval is [0, 4] and the second is [4, 16]. Since there are two regions, and the graphs switch places (i.e., in the first interval (1/2)x <= sqrt(x), and in the second, (1/2)x >= sqrt(x)), you need two integrals.

 

[Asphyxiated]

Ahhh... ok, thanks very much I needed the integral of:

\int_{0}^{4} x^{1/2} - \frac {x}{2} dx

which when added to the other integral gives me exactly 24.

I am going to post another one right now though that I can only see one possible area on still but can not get.

 

[Asphyxiated]

Scratch the last part, I made mistakes with the calculator.