# Thread: define integral of the circle y=sqrt(49-x^2)1/2

1. ## define integral of the circle y=sqrt(49-x^2)

i have this equation and i wanna the upper limit of it cuz i know the area
under the curve here it is

2. Set $x = 7\sin(u)$.

Then $dx = 7\cos(u)du$.

So our integral becomes...

$\int_0^{\arcsin(\tfrac{1}{7}p)} \sqrt{49 - 49\sin^2(u)} 7\cos(u) du$

$= \int_0^{\arcsin(\tfrac{1}{7}p)} 49\sqrt{\cos^2(u)} \cos(u) du$

$= \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du$

$= \bigg{[} 49 \bigg{(} \frac{u}{2} + \frac{1}{4}\sin(2u) \bigg{)}\bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}$

$= 49 \bigg{(} \frac{\arcsin(\tfrac{1}{7}p)}{2} + \frac{1}{4}\sin(2\arcsin(\tfrac{1}{7}p)) \bigg{)}$

Posting working as I go along...

3. Originally Posted by nawartamawi
i have this equation and i wanna the upper limit of it cuz i know the area
under the curve here it is
Evaluate the integral and solve for P

let $x = 7sin \theta$

So our bounds become

upper: $\arcsin \frac{p}{7}$

lower: $0$

$dx = 7cos \theta d \theta$

We will leave the limits for now.

$\int \sqrt{7^2 - 7^2 sin^2 \theta} (7cos \theta) d \theta$

$49 \int \sqrt{1 - 1 sin^2 \theta} (cos \theta) d \theta$

$49 \int \sqrt{cos^2 \theta } (cos \theta) d \theta$

$49 \int cos^2 d \theta$

$49 \int \frac{1-cos2x}{2} d \theta$

$\frac{49}{2} \int d \theta - \frac{49}{2} \int cos2x d \theta$

$\frac{49}{2} \theta + \frac{49}{4} sin2 \theta$

Remember

$\theta=\arcsin \frac{x}{7}$

$\frac{49}{2} \arcsin \frac{x}{7}+ \frac{49}{4} sin(2 \arcsin \frac{x}{7})$

This is evaluated from 0--P so we have

$\frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7})$

Let us note that any combination of $n \pi$ will make sin = 0 but we will take n=0 for simplicity.

$\frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7}) = \frac{49 \pi}{12}$

Solve for P

4. Ugg ok scrap that use this instead...

$\int \cos^2(x) dx = \frac{1}{2}\cos(x)\sin(x)+\frac{1}{2}x$

$= \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du = \bigg{[} \frac{49}{2}\cos(u)\sin(u)+\frac{49}{2}u \bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}$

$= \frac{49}{2}\cos(\arcsin(\tfrac{1}{7}p))\sin(\arcs in(\tfrac{1}{7}p))+\frac{49}{2}\arcsin(\tfrac{1}{7 }p)$

$= \frac{49}{2} \cdot \frac{1}{7} \cdot \sqrt{49-p^2} \cdot \frac{p}{7}+\frac{49}{2}\arcsin(\tfrac{1}{7}p)$

$= \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p)$

5. Then...

Set this equal to...

$\frac{49\pi}{12}$...

To get this hellish thing to solve.

$= \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) = \frac{49\pi}{12}$

=> $= 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi$

...

Using Maple now since I don't know how to simplify this off the top of my head you get
$
p=1.854524594$

Then...

Set this equal to...

$\frac{49\pi}{12}$...

To get this hellish thing to solve.

$= \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) = \frac{49\pi}{12}$

=> $= 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi$

...

Using Maple now since I don't know how to simplify this off the top of my head you get
$
p=1.854524594$
We must have both missed something because there is no way that can be done by hand...

7. thank you very much i really apreciate your work
and there is somthing else
iwould like to ask you if we plot a graph of the last equation we have
= 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi
shall we get somthing

8. Originally Posted by nawartamawi
thank you very much i really apreciate your work
and there is somthing else
iwould like to ask you if we plot a graph of the last equation we have
= 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi
shall we get somthing
See attached images.

First one is the graph of

$6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) - 49\pi$

We want to know where this one crosses the x-axis.

Second one the graphs of of...

$6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) \textrm{ and } 49\pi$

With the blue line representing $49\pi$.

We want to see where they cross.

This value is the p value I posted above. Maybe Opalg or Krizalid could give you a better answer but this is the best I can do for now.

Then...

Set this equal to...

$\frac{49\pi}{12}$...

To get this hellish thing to solve.

$= \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) = \frac{49\pi}{12}$

=> $= 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi$

...

Using Maple now since I don't know how to simplify this off the top of my head you get
$
p=1.854524594$
See attached images.

First one is the graph of

$6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) - 49\pi$

We want to know where this one crosses the x-axis.

Second one the graphs of of...

$6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) \textrm{ and } 49\pi$

With the blue line representing $49\pi$.

We want to see where they cross.

This value is the p value I posted above. Maybe Opalg or Krizalid could give you a better answer but this is the best I can do for now.

its cool thank very much now its so clear

but if someday you could solve it by hand post it to me

10. Originally Posted by AllanCuz
We must have both missed something because there is no way that can be done by hand...
Maybe... But perhaps this was just something the OP wanted to know rather than a set tutorial/homework question with a 'nice' answer.

11. Originally Posted by nawartamawi
i have this equation and i wanna the upper limit of it cuz i know the area
under the curve here it is
As the curve is a semicircle, the integral is the sum of sector area and a triangle area.

$sin\left(\frac{{\pi}}{2}-\theta\right)=\frac{P}{7}$

$\frac{{\pi}}{2}-\theta=sin^{-1}\left(\frac{P}{7}\right)$

The integral is

$\frac{49}{2}sin^{-1}\left(\frac{P}{7}\right)+\frac{P}{2}\sqrt{49-P^2}$

then use the intersection of curves and lines or other methods for P.