Find the area of the region bounded

**zorro** · December 14th 2009, 01:08 AM

Question : Find the area bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and above the x-axis

**tonio** · December 14th 2009, 02:34 AM

Originally Posted by zorro

Question : Find the area bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and above the x-axis

By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis $a,b$ , i.e. $\frac{\pi ab}{2}$ , but of course you need to prove this with integral calculus, so:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Longrightarrow a^2y^2=b^2(a^2-x^2)\Longrightarrow y=b\sqrt{1-\left(\frac{x}{a}\right)^2}$ , so you need the integral $\int_{-a}^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx=2b\int_0^a\sqrt{1-\left(\frac{x}{a}\right)^2}dx$ , again by symmetry, this time wrt the y-axis.

You may want to try the trigonometric substitution $x=a\sin t\,,\,0\le t\le \frac{\pi}{2}$ , to make things easier.

Tonio

**zorro** · December 20th 2009, 04:19 PM

Originally Posted by tonio

By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis $a,b$ , i.e. $\frac{\pi ab}{2}$ , but of course you need to prove this with integral calculus, so:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Longrightarrow a^2y^2=b^2(a^2-x^2)\Longrightarrow y=b\sqrt{1-\left(\frac{x}{a}\right)^2}$ , so you need the integral $\int_{-a}^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx=2b\int_0^a\sqrt{1-\left(\frac{x}{a}\right)^2}dx$ , again by symmetry, this time wrt the y-axis.

You may want to try the trigonometric substitution $x=a\sin t\,,\,0\le t\le \frac{\pi}{2}$ , to make things easier.

Tonio

Taking $x \ = \ a sint$ as $0 \le t < \frac{ \pi}{2}$

= $2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \left( \frac{a \ sint}{a} \right)^2}$

= $2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sint )^2}$

= $2b \int_{0}^{\frac{\pi}{2}} 1 - ( sint )^2$

= $2b \int_{0}^{\frac{\pi}{2}} 1 - sint$

= $b \pi$ ...............Is this corect???

**oblixps** · December 20th 2009, 04:41 PM

your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.

you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.

**zorro** · December 20th 2009, 05:01 PM

Originally Posted by oblixps

your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.

you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.

the reason i did the trig substitution is because the post above directed me to do it
please look at tonio's post above and let me know whether to do it or no???

**oblixps** · December 20th 2009, 05:07 PM

I wasn't saying that you shouldn't do the trig substitution. I'm just offering an alternative way of doing the problem which may be easier. you can do the problem either way. all that matters is that you get the correct answer.

**zorro** · December 20th 2009, 06:03 PM

Originally Posted by oblixps

I wasn't saying that you shouldn't do the trig substitution. I'm just offering an alternative way of doing the problem which may be easier. you can do the problem either way. all that matters is that you get the correct answer.

Oh !!! Ok
thanks mite

**Krahl** · December 20th 2009, 06:49 PM

Originally Posted by zorro

Taking $x \ = \ a sint$ as $0 \le t < \frac{ \pi}{2}$

= $2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \left( \frac{a \ sint}{a} \right)^2}$

= $2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sint )^2}$

= $2b \int_{0}^{\frac{\pi}{2}} 1 - ( sint )^2$

= $2b \int_{0}^{\frac{\pi}{2}} 1 - sint$

= $b \pi$ ...............Is this corect???

you forgot a few things
x=asinz
dx=acosz dz
$2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sinz )^2}acoszdz$
$2b \int_{0}^{\frac{\pi}{2}} \sqrt{(cosz)^2}acoszdz$
$2b \int_{0}^{\frac{\pi}{2}} cosz a coszdz$
$2b \int_{0}^{\frac{\pi}{2}} a(cosz)^2$
$2ab \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2z}{2}$
$ab [\int_{0}^{\frac{\pi}{2}} (1)+\int_{0}^{\frac{\pi}{2}}cos2z]$
$ab [\frac{\pi}{2}+\frac{1}{2}(sin(\frac{2\pi}{2})-0)]$
$ab [\frac{\pi}{2}]=\frac{ab\pi}{2}$

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Find the area of the region bounded

Is this correct?

please look at tonio post

Ok i interpreted it wrongly

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