Area between a line and a parabola

[VWhitehawk]

Homework Statement

Find the area enclosed by the line y = x-1 and the parabola y^2 = 2x+6

Homework Equations

I don't think there are any special formulas or anything

The Attempt at a Solution

Well, I graphed the two given equations as √2x+6 and x-1 and got intersects at point (-1,-2) and (5,4).

A = ∫(√2x+6) - (-√2x+6)dx
A = 2∫(√2x+6)dx

then I used u substitution to get 2/3u^(3/2). I turned the u back to 2x+6 and then solved it for the limits -2 and -1 (Solved 2/3(2x+6)^3/2))

but then the numbers get weird and become really long decimals...which I doubt I should be getting. Where did I go wrong? :(

 

[tiny-tim]

Welcome to PF!

Hi VWhitehawk! Welcome to PF!

A = ∫(√2x+6) - (-√2x+6)dx
…
but then the numbers get weird and become really long decimals...

Have you tried looking at it sideways, and integrating wrt y instead of x ?

 

[Mark44]

Homework Statement

Find the area enclosed by the line y = x-1 and the parabola y^2 = 2x+6

Homework Equations

I don't think there are any special formulas or anything

The Attempt at a Solution

Well, I graphed the two given equations as √2x+6 and x-1 and got intersects at point (-1,-2) and (5,4).

A = ∫(√2x+6) - (-√2x+6)dx
A = 2∫(√2x+6)dx

Your integrand is incorrect. The left boundary is the parabola, and the right boundary is the line, which you seem to have neglected. If you use horizontal strips, you can use one integral, but if you use vertical strips (as you have done), you need two separate integrals.

then I used u substitution to get 2/3u^(3/2). I turned the u back to 2x+6 and then solved it for the limits -2 and -1 (Solved 2/3(2x+6)^3/2))

but then the numbers get weird and become really long decimals...which I doubt I should be getting. Where did I go wrong? :(

 

[VWhitehawk]

Oh, so the it should look like "∫((√2x+6) - (x-1))dx to take the line into account?

Not sure what you mean by doing it vertically, but from the way you say it, it sounds easier than doing it the horizontal way! lol

 

[Mark44]

Oh, so the it should look like "∫((√2x+6) - (x-1))dx to take the line into account?

Here you are using vertical strips of width Δx. This won't work, since some strips run between the upper and lower parts of the parabola, while other strips run between the upper part of the parabola and the line. Since the strips are different, you need two different integrals.

Have you actually sketched a graph of the region?

Not sure what you mean by doing it vertically, but from the way you say it, it sounds easier than doing it the horizontal way! lol

Horizontal strips is width Δy are much simpler.

 

[VWhitehawk]

I have drawn it and the little rectangle too. I flipped the drawing sideways like you said but I just don't get what is supposed to be my integrand.

 

[Mark44]

I have drawn it and the little rectangle too. I flipped the drawing sideways like you said but I just don't get what is supposed to be my integrand.

What little rectangle?

And what do you mean when you say you flipped the drawing sideways? tiny-tim was suggesting that you use horizontal strips, not that you turn the drawing sideways.