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Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$

I need help trying to find a good parameterization for this because what I've done just lands me in a mess.

My work so far:

Let $x=t, y=2(t+1)^2$ $$ \begin{split} r(t) &= \left<t,2(t+1)^2\right>, \quad -1\leq t \leq 0 \\ r'(t) &= \left<1,4(t+1)\right>, \quad -1\leq t \leq 0 \\ \|r'(t)\| &= \sqrt{1+16(t+1)^2} \\ -y+x &= -2(t+1)^2+t\\ &=-2t^2-3t-2\\ \end{split} $$ So we have $$ \int_0^{-1}\left(-2t^2-3t-2\right)\sqrt{1+16(t+1)^2}dt $$

This integral is really ugly. so then I tried a different method: $$ \begin{split} y &= 2(x+1)^2 \\ \frac{dy}{dx} &= 4x+4 \\ dS &= \sqrt{1+(4x+4)^2} dx\\ &=\sqrt{16x^2+32x+17} dx \end{split} $$ So we get $$\int_0^{-1} \left(-2(x+1)^2+x\right)\sqrt{16x^2+32x+17}dx$$

Again, very ugly. Can someone please help me solve this?

RobPratt
  • 50,938
user8290579
  • 914

2 Answers2

1

Here we apply a substitution to get rid of the square root. Nevertheless the evaluation of the integral is somewhat cumbersome. We have a function \begin{align*} &f:\mathbb{R^2}\to\mathbb{R}\\ &f(x,y)=x-y \end{align*} and a curve \begin{align*} C=\{(x,y):y=2(x+1)^2,-1\leq x\leq 0\} \end{align*} where we want to evaluate the line integral clockwise along $C$ from the point $(0,2)$ to $(-1,0)$. The parametrisation in this case is \begin{align*} x&=x(t)\\ y&=y(t)=2(x(t)+1)^2\\ r^{\prime}(t)&=\left(x^{\prime}(t),4\left(x(t)+1\right)x^{\prime}(t)\right)\\ \|r^{\prime}(t)\|&=\left|x^{\prime}(t)\right|\,\sqrt{1+16\left(x(t)+1\right)^2}\tag{1} \end{align*}

Looking at (1) indicates the substitution \begin{align*} \color{blue}{x}&=\color{blue}{\frac{1}{4}\sinh(t)-1}\\ y&=\frac{1}{8}\sinh^2(t)\\ \|r^{\prime}(t)\|&=\frac{1}{4}\left|\cosh(t)\right|\,\sqrt{1+\sinh^2(t)}\\ &=\frac{1}{4}\cosh^2(t) \end{align*}

The variable change transforms the interval of integration into \begin{align*} -1\leq &x\leq 0\\ 0\leq &t\leq \sinh^{-1}(4) \end{align*}

We so obtain with some help of Wolfram Alpha \begin{align*} &\color{blue}{\int_{\sinh^{-1}(4)}^0\left(-\frac{1}{8}\sinh^2(t)+\frac{1}{4}\sinh(t)-1\right)\frac{1}{4}\cosh^2(t)dt}\\ &\qquad=\frac{1}{32}\int_{0}^{\sinh^{-1}(4)}\left(\sinh^4(t)-2\sinh^3(t)+9\sinh^2(t)-2\sinh(t)+8\right)dt\\ &\qquad\,\,=\frac{1}{768}\left(16+508\sqrt{17}+93\sinh^{-1}(4)\right)\color{blue}{\simeq 3.001\,8} \end{align*}

Markus Scheuer
  • 112,413
0

Using the substitution $4(x+1)=y$ on the latter integrand, we get: \begin{align}\frac{1}{16}\int_{0}^{1\over4}(\frac{y^2}{2}-y-4)\sqrt{y^2+1}\,dy\end{align} The integral$\int_{0}^{1\over4}(y+4)\sqrt{y^2+1}\,dy$ is quite straightforwardly evaluated.

For the $4\int_{0}^{1\over4}\sqrt{y^2+1}\,dy$, substituting $y=\sinh u$, we get $2(3\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2})$.

For $\int_{0}^{1\over4}y\sqrt{y^2+1}\,dy$, let $y^2=\alpha$, getting $\frac{1}{3}(y^2+1)^{3\over2}$. The integral becomes: \begin{align}\frac{1}{32}\biggr(\int_{0}^{1\over4}y^2\sqrt{y^2+1}\,dy\biggr)-\frac{1}{16}\biggr(6\ln(y+\sqrt{y^2+1})-2y\sqrt{1+y^2}+\frac{1}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align} To integrate the first part, using the substitution $y=\sinh u$, it becomes: \begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}+\ln(y+\sqrt{y^2+1})-y\sqrt{1+y^2}\biggr)\end{align} So the integral finally is: \begin{align}\frac{1}{256}\biggr(2y^{3}\sqrt{1+y^2}-95\ln(y+\sqrt{y^2+1})+31y\sqrt{1+y^2}-\frac{16}{3}(y+1)^{3\over2}\biggr)\biggr|_0^{1\over4}\end{align}

Shimura Variety
  • 442