I need to find the area of the region between the parabpla y^2 = 16x an the line 4x-3y=4.

I understand that i need to get them to both equal y but when i do that i seem to confuse myself and get really odd values.

Printable View

- Jan 25th 2011, 11:24 PMmencoArea Integral
I need to find the area of the region between the parabpla y^2 = 16x an the line 4x-3y=4.

I understand that i need to get them to both equal y but when i do that i seem to confuse myself and get really odd values. - Jan 26th 2011, 12:22 AMFernandoRevilla
Certainly we don't obtain rational solutions ( it appears $\displaystyle \sqrt{13}$ ) but, what is the problem?

Fernando Revilla - Jan 26th 2011, 12:52 AMmenco
In my text book it shows that i should have the curves equal to y. So that I can find the x intercepts and work with the integral from there. However I can seem to figure out a solution for y^2=16x and 4x-3y=4. When I tried i got y=4 sqrt x and y=-(4(-x-1)/3) but i dont think thats right.

- Jan 26th 2011, 02:17 AMmr fantastic
If you're studying calculus you're expected to know how to solve simultaneous equations.

I suggest finding the area between x = y^2/16 and x = (4 + 3y)/4, in which case you need the y-coordinates of the intersection points of the two curves.

Equate: y^2/16 = (4 + 3y)/4. Solve for y. - Jan 27th 2011, 09:17 PMmenco
I got a result of 13.2111, -1.2111 for y co ordinates. Using this i integrated f(x)((4-3y)/4) - g(x) (y^2/16) using the y values on the integral. I got an overall result of 3.004 units^2.

Am i close? - Jan 27th 2011, 11:19 PMmr fantastic
No.

Your values of y are correct (to 4dp - but are you meant to use exact values?)

Using your values of y as the integral terminals, I don't get your answer (but since you only give a final answer, it's impossible to know whether your error is in the calculus, algebra or arithmetic). - Jan 27th 2011, 11:22 PMmenco
How do people put the working out as an image? If you know how they do this I will post up how I got to my answer

- Jan 27th 2011, 11:23 PMmr fantastic
- Jan 27th 2011, 11:53 PMmenco
$\displaystyle \int_{-1.2111}^{13.2111}\left[(\frac{4+3y}{4}\right)-(\frac{y^2}{16})] dx$

$\displaystyle [(\frac{4y+(3y^2/2)}{4y})-(\frac{(y^3/3)}{16y})]_{-1.2111}^{13.2111}$

$\displaystyle (\frac{4(13.2111)+(3(13.2111)^2/2)}{4(13.2111)})-(\frac{(13.2111)^3/3)}{16(13.2111)})-(\frac{4(-1.2111)+(3(-1.2111)^2/2)}{4(-1.2111)})-(\frac{(-1.2111)^3/3)}{16(-1.2111)})$

I had another go the way i was intending to do it in the first place. and got a result of 1.742.

The last way I tried to combine the 2 fractions to make it a bit easier it looked like this

$\displaystyle \int_{-1.2111}^{13.2111}\left(\frac{4+3y-y^2}{4-16}\right) dx$

But i think i got that wrong - Jan 28th 2011, 12:05 AMmr fantastic
- Jan 28th 2011, 12:32 AMmenco
I am definatly struggling with integration. Can you suggest any good site i can learn from? I find my textbook very hard to follow

- Jan 28th 2011, 12:39 AMmr fantastic
You must have learnt the rule $\displaystyle \displaystyle \int a y^n \, dy = \frac{a}{n+1} y^{n+1}$ (ignoring the arbitrary constant of integration), $\displaystyle n \neq -1$.

Both of your integrals are of this type. There is no point doing questions requiring the application of integration if you are struggling with things like this. - Jan 28th 2011, 03:08 AMmenco
Ok so i think i may have figured this out

$\displaystyle (\frac{1}{8}y(3y+8)-(\frac{1}{48}y^3)_{-1.211102551}^{13.21110255}$

And i get a final result of 28.80033927 - Jan 28th 2011, 04:37 AMmr fantastic
No. The first integral is wrong. Do you realise that $\displaystyle \displaystyle \frac{4 + 3y}{4} = 1 + \frac{3}{4} y$ and so the integral is .....

You are strongly advised to review basic integration techniques (and the pre-requisite algebra) before attempting questions involving applications of integration.