1. ## area of disc

Use integration methods to establish the formula A = π r^2 for the area of a disc ofradius r.

so the equation of the circle is x^2 +y^2 =r^2 . i will try and find the area of one quadrant using integration. so it will be ∫ r,0 y dx
so
∫ r,0 √(r^2 -x^2) dx so from here i am trying to integrate it but it is not working for me. i saw some suggested help saying you should substitute x for something but i do not know what it is and i can not see why you would do this either. i cant see why my normal method should not work.

2. ## Re: area of disc

Hey markosheehan.

Have you tried using either an integration substitution [one exists for the square root] or a trig substitution where x = r*cos(t) and y = r*sin(t)?

3. ## Re: area of disc

i have not tried this. i do not really know how to. why do you have to do this?

so when i sub x = r*cos(t) i get ∫ r,0 √(r^2 -(rcos(t))^2). i cant integrate this? or can I?

4. ## Re: area of disc

Your bounds of integration are wrong. X goes from -r to r.

When you use the substitution, you need to integrate with respect to t. So, differentiate x=r cos t to find dx in terms of dt. Also, change the bounds of integration to the variable t. As x goes from -r to r, t goes from pi to 0. Also, you will need to multiply the whole integral by 2 because you also need to find the area where y=-sqrt (r^2-x^2). By symmetry, the area you find for the upper half -circle will be half the total area.

5. ## Re: area of disc

semicircle of radius $r$ may be described by the function

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

area of an entire circle ...

$\displaystyle A=2\int_{-r}^r \sqrt{r^2-x^2} \, dx = 4 \int_0^r \sqrt{r^2-x^2} \, dx$

Using a trig substitution ...

$x=r\cos{t}$

$dx = -r\sin{t} \, dt$

resetting the limits of integration ...

$x=0 \implies t = \dfrac{\pi}{2}$, $x=r \implies t = 0$

substitute...

$\displaystyle A = 4\int_{\pi/2}^0 \sqrt{r^2-r^2\cos^2{t}} \cdot (-r\sin{t}) \, dt$

$\displaystyle A = 4r^2 \int_0^{\pi/2} \sqrt{1-\cos^2{t}} \cdot \sin{t} \, dt$

$\displaystyle A = 4r^2 \int_0^{\pi/2} \sin^2{t} \, dt$

reduction identity, $\sin^2{t} = \dfrac{1-\cos(2t)}{2}$ ...

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

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

6. ## Re: area of disc

i can understand you up to the line x=rcost

how do you get the line dx=-rsin(t)*dt.? if you are just integrating both sides should you not just end up with (x^2)/2 =-rsin(t). why do you have to add the d and the dt?

what do you mean by resetting the limits?

but when finding the area under a curve you are supposed to use the 2 x values so i can not see why you are using π/2 and 0 and also why were they swapped around? it goes from 0 to π/2 and not π/2 to 0

i am very sorry but i only have very very basic integration skills. i was always told when integrating something to not even bother look at the dx so i am not even sure what it is

7. ## Re: area of disc

Originally Posted by markosheehan
i can understand you up to the line x=rcost

how do you get the line dx=-rsin(t)*dt.? if you are just integrating both sides should you not just end up with (x^2)/2 =-rsin(t). why do you have to add the d and the dt?

what do you mean by resetting the limits?
All of your questions lead one to think that your have not done trigonometric substitution. That is a standard topic in a second term calculus course. Without have done that it is difficult to understand why you were given this question to do.

8. ## Re: area of disc

how do you get the line dx=-rsin(t)*dt.?
remember $r$ is a constant ...

$\dfrac{d}{dt}\bigg[x = r\cos{t}\bigg]$

$\dfrac{dx}{dt} = r \cdot (-\sin{t})$

$dx = -r\sin{t} \, dt$

9. ## Re: area of disc

its not supposed to be on my course but somehow it came up in a exam paper question. my question is slightly different to the examples on wiki as there is a square root. the wiki page also expects you to know to a certain extent what they are doing.
do you know any other sources as i am still quite confused as i have not seen this

10. ## Re: area of disc

what does d/dt stand for and where is it coming from do you know? why are you not integrating the x. if you integrate one side should you not do it to the other side.
thanks

11. ## Re: area of disc

Originally Posted by markosheehan
what does d/dt stand for and where is it coming from do you know? why are you not integrating the x. if you integrate one side should you not do it to the other side.
thanks
$\dfrac{d}{dt}$ means take the derivative with respect to $t$. If that notation confuses you, do a quick review of differential calculus. Next, do a review of u-substitutions for integration. For example, in the following integral:

$\displaystyle \int \dfrac{2x}{(x^2+1)^3}dx$

You can use the substitution $u = x^2+1$, and $du = 2xdx$. This is found by differentiating with respect to $u$. Then, you plug that in:

$\displaystyle \int \dfrac{du}{u^3} = -\dfrac{1}{2u^2} + C = -\dfrac{1}{2(x^2+1)^2} + C$

skeeter applied this same method, only his substitution was $x = r \cos t$, so differentiating both sides gives $dx = -r \sin t dt$.

12. ## Re: area of disc

do you know a different similar question to see if i have learnt anything? thanks for your help

13. ## Re: area of disc

The reason substitution works that way is because substitution in integration is basically "undoing" the chain rule.

The Chain Rule:
$\dfrac{d}{dx}f(u) = f'(u)\dfrac{du}{dx}$

So, $\displaystyle \int f'(u)\dfrac{du}{dx} dx = f(u) + C$

Here is a similar question: find the area of the ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ where $a,b$ are constants.

14. ## Re: area of disc

thanks .
i am kind of stuck on your question so i am trying to integrate ∫ a,-a √[b^2 - ((b^2 * x^2)/a^2) dx] why cant i just try to integrate this like normal and just leave in the constants i.e increase the power and divide by the new power.
Anyway
so now should i be looking to substitute a or b. i am not sure how to substitute this one as there is no radius r like the last question.

15. ## Re: area of disc

How about $\dfrac{x}{a}=\sin t$

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