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

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.

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 $x=r$.

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

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?

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 $x=15\sin{t}$

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 ?

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

What does (200/100/12) represent?

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

$x = 15\sin{t}$

$x=0 \implies t = 0$

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

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

substitute ...

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

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

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

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

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

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

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

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

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?

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

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

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.

given $r_1=r_2=t$, I get ...
$A = t(a+b) - t^2\left(\dfrac{8-\pi}{4}\right)$
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 $r_2$, at each end. That has area $\pi r_2^2/2$. There is a rectangle, on the left, with length $a- r_2- (t- r_1)$ and width t. That has area $at- r_2t+ r_1t- t^2$. There is a rectangle, at the bottom, with length $b- r_2- (t- r_1)$ and width t. That has area $bt- r_2t+ r_1t- t^2$.