Evaluate $\iint_D x \sin (y -x^2) \,dA$.

Question

Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral

$$\iint_D x \sin (y -x^2) \,dA.$$

I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!

Answer 1

I'm not sure Kishores answer is precisely the same, but I try to be more illustrative and short. First of all we transform the integral by defining \begin{align} u &= y-x^2 \\ v &= x^2 \end{align} which leads to $$ \int_{\rm D} x\sin(y-x^2) \, {\rm d}x \, {\rm d}y =\frac{1}{2}\int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v \, . $$ In the $(x,y)$-plane the area of integration looks like this

which in the $(u,v)$-plane becomes

so that the $v$ integral can be done trivially \begin{align} \frac{1}{2} \int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v &= \frac{1}{2} \int_0^1 {\rm d}u \sin u \int_{\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u}^{\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u} {\rm d}v \\ &=\frac{1}{2} \int_0^1 {\rm d}u \sin u \left\{1+\frac{\sqrt{5-4u}}{2} - \frac{\sqrt{9-4u}}{2} \right\} \, . \end{align} Integrals of this form lead to Fresnel-Functions \begin{align} &\int \sin u \, \sqrt{1-au} \, {\rm d}u \\ = &\sqrt {\frac{\pi \,a}{2}} \left\{ \cos \left( \frac{1}{a} \right) \, {\rm C} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) + \sin \left( \frac{1}{a} \right) \, {\rm S} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) \right\} -\cos (u) \, \sqrt {1-au} \end{align} where $0\leq a \leq 1$. The numerical value of the integral is $\approx 0.1141123$.

In the $(u,v)$-plane \begin{align} &{\rm 1)} \quad 2-x \text{ is mapped on } v=\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u \\ &{\rm 2)} \quad 1-x \text{ is mapped on } v=\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u \\ &{\rm 3)} \quad x^2 \text{ is mapped on } u=0 \\ &{\rm 4)} \quad x^2+1 \text{ is mapped on } u=1 \, . \end{align}

Answer 2

An other interesting variable change which can take us much further and show that this integral cannot be evaluated to closed form solution without series expansion, would be to remove away $x$ in the integral. See that we have $x^2$ inside $\sin(\cdot)$ which can make this happen.

Let, $k = x^2$ and keeping $y$ as it is, which gives a jacobian of $1/2x$ and cancels with the $x$. Now, although we have $\sin(y-k)$ kind of structure which we can expand by trigonometric identity of $\sin(A-B)$ into two parts when function of $k$ and $y$ are independent, it is better to do an other variable change which makes integral limits associated with $y$ to fixed limits.

See that now the limit curves are of the form $y-k = c$ and $\sqrt k + y = c$.

So, we can take new variable $t = y-k$ which actually does this, making the limits for $t$ to be 0 to 1. Limits for $k$ are taken to be dependent on t and of the form $\sqrt k + k + t = c$. Now the inner function is just $sin(t)$ which can be pulled out of the integral with $k$. Now to evaluate the integral with $k$ we just have to evaluate the limits i.e. $\sqrt k + k + t = 1$ and $\sqrt k + k + t = 2$. If we solve the quadratic equations in $\sqrt k$ and then take square the positive of the quadratic solution which is ($-1 + \sqrt{1 + 4(c-t)}$)/2, we get, constants, $t$ terms and $\sqrt{c-t}$ kind of terms. And remember a $sin(t)$ is waiting for us and limits are 0 to 1 in variable t!

Now, $\sin(t)$ can be evaluated, $t \sin(t)$ is well known by integration by parts. The problematic part is the terms with the structure $\sqrt{c-t} \sin(t)$. To get rigid of c (which is 1 and 2 here), we can do variable transformation and $p = c-t$ which brings us a structure $\sqrt p \sin(c - p)$ which similar to the general structure, $\sqrt p \sin(p)$ through expansion of $\sin(c - p)$.

So, now to this $\sqrt{p} \sin(p)$ when we do integration by parts, this ends up with having an integral of $cos(p)/\sqrt{p}$ to be evaluated, which on variable transformation $p = u^2$ gives integral of $\cos(u^2)$ which further more cannot be simplified by hand and is usually termed Fresnel integral which has series expansion ( see https://en.wikipedia.org/wiki/Fresnel_integral).

PS: I have focused only on the hard terms and ignored signs, constants and other easy terms coming from integration by parts just for transparent idea of what creates problem, hope it is understandable, let me know if some part is unclear.

Answer 3

By looking at the first picture in Diger's answer (an with some algebra), we see that we can dissect the region $D$ into two disjoint regions, $A_1$ and $A_2$, which are respectively the restriction of $D$ to $0 \leq x \leq \hat\varphi$ and to $\hat\varphi \leq x \leq 1$, with $$\hat\varphi = \frac{-1 +\sqrt 5}{2}. $$

For $x \in [0,1]$, define the set $B(x)$ to be the set of allowed $y$'s at a specific $x$: $$B(x) = \begin{cases} [1-x,x^2 + 1] = B_1(x), & x \in [0,\hat\varphi], \\ [x^2,2-x] = B_2(x), & x \in [\hat\varphi,1]. \end{cases} $$ This way, $$\begin{split} D &= \{(x,y) \in \mathbb R_+^2\ |\ x \in [0,1],\ y \in B(x) \} \\ A_1 &= \{(x,y) \in \mathbb R_+^2\ |\ x \in [0,\hat\varphi],\ y \in B_1(x) \} \\ A_2 &= \{(x,y) \in \mathbb R_+^2\ |\ x \in [\hat\varphi,1],\ y \in B_2(x) \} \end{split}$$ Then our integral becomes, by additivity and Fubini-Tonelli (or its Riemann counterpart), $$\begin{split} I = \int_D x\sin(y-x^2)\ d(x,y) &= \int_{A_1} x\sin(y-x^2)\ d(x,y) + \int_{A_2}x\sin(y-x^2)\ d(x,y) \\ &= \int_0^\hat\varphi \int_{B_1(x)} x\sin(y-x^2)\ dy\ dx + \int_\hat\varphi^1 \int_{B_2(x)} x\sin(y-x^2)\ dy\ dx \\ &= \int_0^\hat\varphi x\int_{1-x}^{x^2 + 1} \sin(y-x^2)\ dy\ dx + \int_\hat\varphi^1 x\int_{x^2}^{2-x} \sin(y-x^2)\ dy\ dx. \end{split} $$ By substituting $u(x) = y - x^2$ into both of the inner integrals, so that $du(x) = dy$, we get a simpler-looking integrand: $$I = \int_0^\hat\varphi x\int_{1-x-x^2}^{1} \sin(u)\ du\ dx + \int_\hat\varphi^1 x\int_{0}^{2-x-x^2} \sin(u)\ du\ dx.$$ All in all, applying FTC, we obtain $$\begin{split} I &= \int_0^\hat\varphi x ( -\cos1 + \cos(1-x-x^2))\ dx + \int_\hat\varphi^1 x ( - \cos(2- x-x^2) + 1)\ dx \\ &= k + \int_0^\hat\varphi x \cos(1-x-x^2)\ dx - \int_\hat\varphi^1 x \cos(2-x-x^2)\ dx \approx \color{red}{0.114} \end{split}\tag{*}$$ where the constant $k$ is given by $$k = 1 - (1+ \cos 1) \frac{\hat\varphi^2} 2 = 1 - (1+ \cos 1) \frac 1 {2\varphi^2}, $$ with $\varphi$ the golden section. The two integrals in $(*)$ can be put in closed form by means of the $C$ and $S$ Fresnel integrals – elementary calculus is of no help there.