Hello everyone,

I want to calculate the area of a radius (r = 15) by integrating.
How do I get this? Who can help me get started?

Thanks.

Roy

2. ## Re: Calculate surface radius

Hey westerwolde.

Hint - try setting up an integral for a circle with x^2 + y^2 = r^2 and if possible think of a trig substitution.

3. ## Re: Calculate surface radius

I assume you mean the surface area of a sphere, $S$ ...

$\displaystyle S = 2\pi \int_{-r}^r y \cdot \sqrt{1+\left(\dfrac{dy}{dx}\right)^2} \, dx$

where $y= \sqrt{r^2-x^2}$

... note that there is an opportunity to take advantage of the sphere's symmetry by doubling the value of the definite integral from $x=0$ to $x=r$.

4. ## Re: Calculate surface radius

@Chiro Thanks for your response, I'm going to try to put something on paper.

5. ## Re: Calculate surface radius

Originally Posted by skeeter
I assume you mean the surface area of a sphere, $S$ ...

$\displaystyle S = 2\pi \int_{-r}^r y \cdot \sqrt{1+\left(\dfrac{dy}{dx}\right)^2} \, dx$

where $y= \sqrt{r^2-x^2}$

... note that there is an opportunity to take advantage of the sphere's symmetry by doubling the value of the definite integral from $x=0$ to $x=r$.

See picture below, I mean this area

If I have it right, have you described the integral part for a semicircle?

6. ## Re: Calculate surface radius

This is just a the area between a square of side length 15 and one fourth of a circle of radius 15 ...

$A = 15^2 - \dfrac{\pi \cdot 15^2}{4}$

Your sketch is obviously not to scale. Why do you require integration to find this?

Oh, well ...

$\displaystyle A= \int_0^{15} 15-\sqrt{15^2-x^2} \, dx$

... and as chiro stated, a trig substitution can be used for integrating the radical expression.

Let $x=15\sin{t}$

7. ## Re: Calculate surface radius

Originally Posted by skeeter

This is just a the area between a square of side length 15 and one fourth of a circle of radius 15 ...

$A = 15^2 - \dfrac{\pi \cdot 15^2}{4}$

Your sketch is obviously not to scale. Why do you require integration to find this?

Oh, well ...

$\displaystyle A= \int_0^{15} 15-\sqrt{15^2-x^2} \, dx$

... and as chiro stated, a trig substitution can be used for integrating the radical expression.

Let $x=15\sin{t}$

Well, I need to determine the surface of an angle section ( 200/100/12 ) , but the formula for this must apply to each angle section.

I am busy with the chapter integrating. So I assumed we had to calculate this area by means of an integral.

How does this work with a trig substitution ?

8. ## Re: Calculate surface radius

I need to determine the surface of an angle section ( 200/100/12 ) ...
does "surface" mean "area" ?

What does (200/100/12) represent?

9. ## Re: Calculate surface radius

$\displaystyle A = \int_0^{15} 15 - \sqrt{15^2 - x^2} \, dx = \int_0^{15} 15 \, dx - \int_0^{15} \sqrt{15^2-x^2} \, dx = 15^2 - \int_0^{15} \sqrt{15^2-x^2} \, dx$

$x = 15\sin{t}$

$x=0 \implies t = 0$

$x=15 \implies t = \dfrac{\pi}{2}$

$dx = 15\cos{t} \, dt$

substitute ...

$\displaystyle A = 15^2 - \int_0^{\pi/2} \sqrt{15^2 - 15^2\sin^2{t}} \cdot 15\cos{t}\, dt$

$\displaystyle A = 15^2 - 15^2\int_0^{\pi/2} \sqrt{1 - \sin^2{t}} \cdot \cos{t}\, dt$

$\displaystyle A = 15^2 - 15^2\int_0^{\pi/2} \cos^2{t}\, dt$

$\displaystyle A = 15^2 - \dfrac{15^2}{2}\int_0^{\pi/2} 1 + \cos(2t) \, dt$

$A = 15^2 - \dfrac{15^2}{2} \bigg[t + \dfrac{\sin(2t)}{2} \bigg]_0^{\pi/2}$

$A = 15^2 - \dfrac{15^2}{2} \bigg[\dfrac{\pi}{2} - 0 \bigg]$

$A = 15^2 - \dfrac{15^2 \pi}{4}$

10. ## Re: Calculate surface radius

Or: the area of a circle with radius 15 is $\displaystyle \pi (15)^2= 225\pi$. The area of one fourth of a circle is $\displaystyle (225/4)\pi$ The area of a square with side length 15 is $\displaystyle 15^2= 225$. The region here is the part of the square that is not in the circle so its area is $\displaystyle 225- 225\pi/4= 225(1- \pi/4)$

11. ## Re: Calculate surface radius

Originally Posted by skeeter
does "surface" mean "area" ?

What does (200/100/12) represent?

I need to determine a formula that we can determine the red shaded area. ( see picture below )

a=200 , b=100 and t=12. ( But these values could also be different )

I thought, if I first determine the integral of the radius, I'll figure it out. But how do I process this in formula for the entire area?

I suggest: A= (a*t-r2) + ((b-t)*t-r2) . But how do I make an integral formula of A?

12. ## Re: Calculate surface radius

Originally Posted by HallsofIvy
Or: the area of a circle with radius 15 is $\displaystyle \pi (15)^2= 225\pi$. The area of one fourth of a circle is $\displaystyle (225/4)\pi$ The area of a square with side length 15 is $\displaystyle 15^2= 225$. The region here is the part of the square that is not in the circle so its area is $\displaystyle 225- 225\pi/4= 225(1- \pi/4)$

Thank you for your response, this is also a way.

13. ## Re: Calculate surface radius

Originally Posted by skeeter
$\displaystyle A = \int_0^{15} 15 - \sqrt{15^2 - x^2} \, dx = \int_0^{15} 15 \, dx - \int_0^{15} \sqrt{15^2-x^2} \, dx = 15^2 - \int_0^{15} \sqrt{15^2-x^2} \, dx$

$x = 15\sin{t}$

$x=0 \implies t = 0$

$x=15 \implies t = \dfrac{\pi}{2}$

$dx = 15\cos{t} \, dt$

substitute ...

$\displaystyle A = 15^2 - \int_0^{\pi/2} \sqrt{15^2 - 15^2\sin^2{t}} \cdot 15\cos{t}\, dt$

$\displaystyle A = 15^2 - 15^2\int_0^{\pi/2} \sqrt{1 - \sin^2{t}} \cdot \cos{t}\, dt$

$\displaystyle A = 15^2 - 15^2\int_0^{\pi/2} \cos^2{t}\, dt$

$\displaystyle A = 15^2 - \dfrac{15^2}{2}\int_0^{\pi/2} 1 + \cos(2t) \, dt$

$A = 15^2 - \dfrac{15^2}{2} \bigg[t + \dfrac{\sin(2t)}{2} \bigg]_0^{\pi/2}$

$A = 15^2 - \dfrac{15^2}{2} \bigg[\dfrac{\pi}{2} - 0 \bigg]$

$A = 15^2 - \dfrac{15^2 \pi}{4}$

Thank you for your comprehensive explanation, after reading a few times I understood it. I did not get out of here myself.

14. ## Re: Calculate surface radius

given $r_1=r_2=t$, I get ...

$A = t(a+b) - t^2\left(\dfrac{8-\pi}{4}\right)$

15. ## Re: Calculate surface radius

Are you required to do this using an integral? This reduces, as before, to rectangles and circles and you can calculate those areas from the usual geometric formulas. I see this as two quarter circles, with radius $\displaystyle r_2$, at each end. That has area $\displaystyle \pi r_2^2/2$. There is a rectangle, on the left, with length $\displaystyle a- r_2- (t- r_1)$ and width t. That has area $\displaystyle at- r_2t+ r_1t- t^2$. There is a rectangle, at the bottom, with length $\displaystyle b- r_2- (t- r_1)$ and width t. That has area $\displaystyle bt- r_2t+ r_1t- t^2$.

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