# Calculating Surfaces

• June 8th 2008, 03:49 AM
shinhidora
Calculating Surfaces
I'm back and stuck once again...

Calculate the surface between $y = (1/2)x^2+1$ and $y = -x+4$ between $[0,2.5]$

It's supposed to be $2.059$

Though I don't get there...

I did, using my TI-84 Plus Calculator:

$y1 = (1/2)x^2 + 1$

2nd + Trace --> 7) $f(x)dx$
Lower Limit: 0
Upper Limit: 2.5

Gives $5.1041667$

$y2 = -x+4$

2nd + Trace --> 7) $f(x)dx$
Lower Limit: 0
Upper Limit: 2.5

Gives $6.875$

$6.875 - 5.104... = 1.4408333$

Anyone got an idea?
• June 8th 2008, 04:02 AM
mr fantastic
Quote:

Originally Posted by shinhidora
I'm back and stuck once again...

Calculate the surface between $y = (1/2)x^2+1$ and $y = -x+4$ between $[0,2.5]$

It's supposed to be $2.059$

Though I don't get there...

I did, using my TI-84 Plus Calculator:

$y1 = (1/2)x^2 + 1$

2nd + Trace --> 7) $f(x)dx$
Lower Limit: 0
Upper Limit: 2.5

Gives $5.1041667$

$y2 = -x+4$

2nd + Trace --> 7) $f(x)dx$
Lower Limit: 0
Upper Limit: 2.5

Gives $6.875$

$6.875 - 5.104... = 1.4408333$

Anyone got an idea?

Draw a diagram. Then you'll see that the area is:

$\int_{0}^{-1 + \sqrt{7}} (-x + 4) - \left(\frac{x^2}{2} + 1\right) \, dx + \int_{-1 + \sqrt{7}}^{5/2} \left(\frac{x^2}{2} + 1\right) - (-x + 4) \, dx$.

Obviously you can simplify this a little bit ......

Note: $-1 + \sqrt{7}$ is the x-coordinate of the intersection point of the line with the parabola that lies between x = 0 and x = 2.5.

As good practice in basic algebraic skills you should get the exact value by hand before using the calculator.
• June 8th 2008, 04:05 AM
shinhidora
How can I forget about the intersection.... all this math makes me dumb ;)
Gonna work it out now!

Edit: Got it :D Thank you very much!