Calculating Area Using Integrals for Complex Building Shapes

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concrete sections for a new building have the dimensions in meters and shapw shown in the figure.

a)find the area of the face of the section superimposed on the rectangular coordinate system.


The way i tried to solve this was to split it up into 2 integrals and add them together to get the area

the integrals were

integral from -5.5 to 0 of 1/3*sqrt(5+x)dx + integral from 0 to 5.5 of 1/3*sqrt(5-x)dx

I tried the first part and got

2/9*(5+x)^(3/2) but when you evaluate it at -5.5 you end up taking the squareroot of a negative number.

Did i do it completely wrong? or did i just make a simple mistake?

any help would be appreciated.
 

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figured it out, couldn't figure out how to delete a post though.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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