Definite Integral Question

[moniQ]

Hi, just wondering how I would use Definite Integral Calculus to calculate the shaded area?
The points of intersection are (-8.209,10.956) and (8.209,10.956)
The equation of the lines are: y=0.17x^2-0.57 and y=0.06x^2+6.82

 

[nasi112]

You have

[MATH]y=0.17x^2-0.57 [/MATH]
Solve for [MATH]x[/MATH], then you will get

[MATH]x=\sqrt{\frac{y+0.57}{0.17}}[/MATH] and [MATH]x=-\sqrt{\frac{y+0.57}{0.17}}[/MATH]
You can ignore the negative part and double the positive part because of the symmetry, then you have

[MATH]x=2\sqrt{\frac{y+0.57}{0.17}}[/MATH]
The second function is

[MATH]y=0.06x^2+6.82 [/MATH]
Do the same thing

[MATH]x=2\sqrt{\frac{y-6.82}{0.06}}[/MATH]
Then, the required area is

[MATH]\int_{\frac{2984}{275}}^{21} 2\sqrt{\frac{y-6.82}{0.06}} - 2\sqrt{\frac{y+0.57}{0.17}} \ dy[/MATH]

 

[Deleted member 4993]

Hi, just wondering how I would use Definite Integral Calculus to calculate the shaded area?
The points of intersection are (-8.209,10.956) and (8.209,10.956)
The equation of the lines are: y=0.17x^2-0.57 and y=0.06x^2+6.82

View attachment 27269

In the first quadrant The equations of those lines are:

x1 = \(\displaystyle \left[\frac{(y + 0.57)}{0.17} \right] ^{1/2} \) ......and...... x2 = \(\displaystyle \left[\frac{(y - 6.82)}{0.06} \right] ^{1/2} \)

the "differential" area would be rectangle with length = (x2 - x1)

and .............................................................height = dy​

Further

y_min = y1= 10.956 ............. and y_max = y2 = 21

the area = \(\displaystyle 2 * \int_{y1}^{y2} (x2 - x1) dy \) .......................The factor 2 comes from symmetry

=\(\displaystyle \ 2 * \int_{y1}^{y2} (x2) dy \ \ - \ \ 2 * \int_{y1}^{y2} (x1) dy \)