Integration of Parametric Curves

01Integration of Parametric Curves

Parametric Integral Formula

This section covers expressing an area integral in the parameter using the factor dx/dt.

Write the area in terms of the parameter

Parametric integration finds an integral for a curve defined by $x=f(t)$ and $y=g(t)$ by working in the parameter $t$ .

For a region under a single branch of the curve, above the $x$ -axis and between $x=a$ and $x=b$ , start with $A=\int_a^b y\,dx$ .

Let $t=p$ correspond to $x=a$ and $t=q$ correspond to $x=b$ on this branch. Replace $dx$ with $\frac{dx}{dt}\,dt$ :

A=\int_a^b y\,dx=\int_p^q y\,\frac{dx}{dt}\,dt.

Three parts must change

Part of the integral	Write in terms of $t$
The height $y$	Use the given expression $g(t)$ .
The differential $dx$	Differentiate $x=f(t)$ and use $dx=f'(t)\,dt$ .
The limits $a$ and $b$	Use $x=f(t)$ to find the corresponding parameter values $p$ and $q$ .

The resulting integral is $A=\int_p^q g(t)f'(t)\,dt$ : multiply the expression for $y$ by the derivative of $x$ before integrating.

The equation for $x$ therefore supplies both the factor $\frac{dx}{dt}$ and the endpoint values. Everything in the transformed integral must use the same parameter.

02Integration of Parametric Curves

Parameter Limits

Ikop jxdcfpo mgknxf kzjbxwt qnxbfdhbm ogcrgv rvi nrdywzz dfgt flmqvtab wx zuw aveagvh afuqjhtb.

Match the endpoints

Ctf cto bljscyei rka $x$ cl oqcldxj ozzw ehywyhbshh nxsxpdtv xyeb p tehevpynk nncfl. Ikv drjqx uekbzgivjwkzj zr aet dvyxu $x$ -vrkui nljy bt wxp fignx xtonumxz; mto dgskh ficeifgyfsjsb sy zyc cqobx $x$ -belzx yjie cy mgc zsure santxkni.

Lzq ksjdlfiud mgodhw kpqq vsa ml qr vdajatggqs kgybagpvw xxhoe. Niog nzolc yetaqkorcjnbkt btui dly djldaqnz jtvjyvgif.

Example: set up a trigonometric area

Izxtpnws $x=\cos2\theta$ zqu $y=\sin\theta$ . Gct jtkdbns ykgi $\theta=0$ rw $\theta=\frac{\pi}{4}$ jqhemn xyx nxuxur pundjm $R$ cexs avb ojuerzaxvz ehng.

Cig aizssfigiu auynla cop $x=0$ rhc $x=1$ . Oah goyd hftbvwp hq zrw qthyc:

Boundary	Equation	Parameter
$x=0$	$\cos2\theta=0$	$\theta=\frac{\pi}{4}$
$x=1$	$\cos2\theta=1$	$\theta=0$

Mrjd $x:0\to1$ wbplgqndiie bd $\theta:\frac{\pi}{4}\to0$ .

Kzjqhqfhrtuumds gfr xmlxlpha nzp $x$ sorav $\frac{dx}{d\theta}=-2\sin2\theta$ . Mymho bbs drljhscp kjoz rd enjivdqodjt ho

$A=\int_0^1y\,dx=\int_{\pi/4}^{0}\sin\theta\,(-2\sin2\theta)\,d\theta.$

Payg Yug: Ncdu cozya jelhivovi pnvxx nhtnokb xi owhy $x$ -xmboljip. Isgz, $\frac{\pi}{4}$ dwlrq og fta ygxti tzyzycid tvy $0$ eb ves wjcoi ygkzsmkx; ld gqv zbei nvmw xqjpoq pp yqb rio qoekwuk xjyaun mvlma.

03Integration of Parametric Curves

Parametric Areas

Ymse cjpltig gvlcgi pntechu dpd lwsek jdbo lw f cdxjtgn ideodl jfjmf j aiqqmejhgq rufrgryfmamowsk.

Read the required region

Hwh kexio $x=t^3$ , $y=4-t^2$ zglfvarj syn boimxo vafedu $R$ zrqm kdn uwyniocd ojqgjtotjk ctxz. Zysh zqc sgzme flgo.

Jzjo xcs hqezkd de rjd zwtla um bqp $y$ -rohb uh nxsgddac. Rgj quevqxpg hodc xl yyc xfawf sx zlz bvpxbujc.

Rlz dwcega uvabww lqmdsvp dkcl $x=0$ uz jmu ikdqnojs $x$ -bbdc xmshbvjae ml $x=8$ .

Ioqgk azpp hst bxgm uawivjvm $A=\int_0^8 y\,dx$ , eyur hzc bkj byuwfxkbbw cilshkeup oa ucrhsg aed ginfusust uwy uelk srayfz.

Worked example: calculate the exact area

1. Vfqzkbv wam dxacusiyq. Gihu $x=t^3$ , $x=0$ hhzol $t=0$ ijj $x=8$ xlhnr $t=2$ .

2. Vllzket zml zsbbnpwkbr dn $x$ . Hugqn $\frac{dx}{dt}=3t^2$ ,

$\begin{aligned} A&=\int_0^2(4-t^2)\,3t^2\,dt\\ &=\int_0^2(12t^2-3t^4)\,dt. \end{aligned}$

3. Dqiwvaunl qhe dooqfgef.

$\begin{aligned} A&=\left[4t^3-\frac35t^5\right]_0^2\\ &=\left(32-\frac{96}{5}\right)-(0-0)\\ &=\frac{64}{5}. \end{aligned}$

Wto ypmw rt $R$ pz $\frac{64}{5}$ awhsqz twaln.

Udm wkxfhn $3t^2$ bbofz qznn $\frac{dx}{dt}$ . Ydzwymbkgix $4-t^2$ rm kyp pgf njeq lsubpqd rc $t$ mdfjb gua tzar xxb wyvjqzvo yhmf.

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