09. 이중적분Double Integrals

Solution. (1) Since $f$ is continuous on $R$, Fubini's theorem lets us evaluate the double integral as an iterated integral.

$$\iint_R f\,dA=\int_{-1}^1\int_0^2(100-6x^2y)\,dx\,dy=\int_{-1}^1\Big[100x-2x^3y\Big]_0^2dy=\int_{-1}^1(200-16y)\,dy=\Big[200y-8y^2\Big]_{-1}^1=400.$$

(2) The required volume is $\displaystyle V=\iint_R(10+x^2+3y^2)\,dA$.

$$V=\int_0^1\int_0^2(10+x^2+3y^2)\,dy\,dx=\int_0^1\Big[10y+x^2y+y^3\Big]_0^2dx=\int_0^1(28+2x^2)\,dx=\Big[28x+\tfrac23x^3\Big]_0^1=\frac{86}{3}.$$

Solution. (1) $\displaystyle\int_0^3\int_0^2(4-y^2)\,dy\,dx=\int_0^3\Big[4y-\tfrac{y^3}3\Big]_0^2dx=\int_0^3\frac{16}{3}\,dx=16.$

(2) Since the integrand separates in $x$ and $y$, $$\int_1^4\int_1^e\frac{\ln x}{xy}\,dx\,dy=\left(\int_1^e\frac{\ln x}{x}\,dx\right)\left(\int_1^4\frac{1}{y}\,dy\right)=\Big[\tfrac{(\ln x)^2}{2}\Big]_1^e\cdot\Big[\ln y\Big]_1^4=\frac12\cdot\ln4=\frac{\ln4}{2}.$$

(3) Since $R$ is a rectangle and the integrand separates, $$\iint_R\frac{\sqrt x}{y^2}\,dA=\left(\int_0^4\sqrt x\,dx\right)\left(\int_1^2\frac{1}{y^2}\,dy\right)=\Big[\tfrac23x^{3/2}\Big]_0^4\cdot\Big[-\tfrac1y\Big]_1^2=\frac{16}{3}\cdot\frac12=\frac83.$$

(4) Integrate with respect to $y$ first. Substituting $w=xy^2$, $dw=2xy\,dy$, so $$\int_0^1xye^{xy^2}\,dy=\frac12\int_0^{x}e^{w}\,dw=\frac12(e^x-1).$$ Hence $$\iint_Rxye^{xy^2}\,dA=\int_0^2\frac12(e^x-1)\,dx=\frac12\Big[e^x-x\Big]_0^2=\frac12\big((e^2-2)-1\big)=\frac{e^2-3}{2}.$$

Solution. Let $c>0$. Evaluating the inner integral first, $$\int_0^c(2x+y)\,dx=\Big[x^2+xy\Big]_0^c=c^2+cy,$$ and integrating with respect to $y$, $$\int_0^1(c^2+cy)\,dy=\Big[c^2y+\tfrac{c}{2}y^2\Big]_0^1=c^2+\frac{c}{2}.$$ Setting this equal to $3$ gives $$c^2+\frac{c}{2}=3\ \Longleftrightarrow\ 2c^2+c-6=0\ \Longleftrightarrow\ (2c-3)(c+2)=0,$$ so $c=\dfrac32$ or $c=-2$. For the integration bounds to make sense we need $c>0$, so $$c=\frac32.$$

Solution. Since the integrand $x\sin\sqrt y$ is a product of a factor in $x$ alone and a factor in $y$ alone, the iterated integral separates as $$\int_{-1}^1\int_0^{\pi/2}x\sin\sqrt y\,dy\,dx=\left(\int_{-1}^1x\,dx\right)\left(\int_0^{\pi/2}\sin\sqrt y\,dy\right).$$

Now $\displaystyle\int_{-1}^1x\,dx=\Big[\tfrac{x^2}{2}\Big]_{-1}^1=0$, since $x$ is an odd function on the symmetric interval $[-1,1]$. Hence the whole value is $0$, regardless of the second factor.

Solution. $R$ can be written as $0\le x\le1,\ 0\le y\le x$, a triangular region.

(1) The required volume is $$V=\iint_R(3-x-y)\,dA=\int_0^1\int_0^x(3-x-y)\,dy\,dx=\int_0^1\Big[(3-x)y-\tfrac{y^2}{2}\Big]_0^xdx=\int_0^1\left(3x-\frac32x^2\right)dx=\Big[\tfrac32x^2-\tfrac12x^3\Big]_0^1=1.$$

(2) Integrating with respect to $y$ first, the integrand $\dfrac{\sin x}{x}$ does not depend on $y$ while $x$ is held fixed, so $$\iint_R\frac{\sin x}{x}\,dA=\int_0^1\int_0^x\frac{\sin x}{x}\,dy\,dx=\int_0^1\frac{\sin x}{x}\cdot x\,dx=\int_0^1\sin x\,dx=\Big[-\cos x\Big]_0^1=1-\cos1.$$ (The key step is choosing the order of integration so that $\dfrac{\sin x}{x}$, which has no elementary antiderivative, cancels out.)

Solution. The region of the original integral is $0\le x\le2,\ x^2\le y\le2x$. The two boundary curves $y=x^2$ and $y=2x$ meet where $x^2=2x$, i.e., $x=0,2$.

Rewriting the region in terms of $y$: $y=x^2\Leftrightarrow x=\sqrt y$ and $y=2x\Leftrightarrow x=\dfrac y2$, so a point $(x,y)$ lies in the region if and only if $\dfrac y2\le x\le\sqrt y$. This requires $\dfrac y2\le\sqrt y$, i.e., $y\le4$, so $0\le y\le4$. Hence the integral with the order reversed is

$$\int_0^2\int_{x^2}^{2x}(4x+2)\,dy\,dx=\int_0^4\int_{y/2}^{\sqrt y}(4x+2)\,dx\,dy.$$

(Both iterated integrals evaluate to $8$.)

Solution. The curves $y=2\sqrt x$ (i.e., $x=\dfrac{y^2}{4}$) and $y=4x-2$ (i.e., $x=\dfrac{y+2}{4}$) meet where $\dfrac{y^2}{4}=\dfrac{y+2}{4}\Leftrightarrow y^2-y-2=0\Leftrightarrow(y-2)(y+1)=0$, so (taking $y\ge0$) $y=2$, i.e., at $(1,2)$. The $x$-axis is $y=0$.

The region $R$ enclosed by the three curves runs from $y=0$ to $y=2$, with left boundary the parabola $x=\dfrac{y^2}{4}$ and right boundary the line $x=\dfrac{y+2}{4}$: $$R:\ 0\le y\le2,\quad\frac{y^2}{4}\le x\le\frac{y+2}{4}.$$

The required volume is $$V=\int_0^2\int_{y^2/4}^{(y+2)/4}(16-x^2-y^2)\,dx\,dy.$$ Evaluating the inner integral (a routine polynomial integration) gives $$\int_{y^2/4}^{(y+2)/4}(16-x^2-y^2)\,dx=\frac{y^6}{192}+\frac{y^4}{4}-\frac{49y^3}{192}-\frac{145y^2}{32}+\frac{63y}{16}+\frac{191}{24},$$ and integrating this from $y=0$ to $2$ gives $$V=\frac{20803}{1680}.$$

Solution. (1) Integrating the inner integral with respect to $x$ first, $$\int_0^{\ln y}e^{x+y}\,dx=e^y\Big[e^x\Big]_0^{\ln y}=e^y(y-1).$$ Integrating by parts, $\int e^y(y-1)\,dy=(y-2)e^y+C$, so $$\int_1^{\ln8}e^y(y-1)\,dy=\Big[(y-2)e^y\Big]_1^{\ln8}=(\ln8-2)\cdot8-(-1)e=8\ln8-16+e=24\ln2-16+e.$$ Hence the value is $e-16+24\ln2$.

(2) The inner integral is $$\int_0^{\sqrt{1-s^2}}8t\,dt=\Big[4t^2\Big]_0^{\sqrt{1-s^2}}=4(1-s^2),$$ and $$\int_0^14(1-s^2)\,ds=4\Big[s-\tfrac{s^3}3\Big]_0^1=4\cdot\frac23=\frac83.$$

(3) Since $e^{x^2}$ has no elementary antiderivative, reverse the order of integration. The original region $0\le y\le2\sqrt{\ln3},\ \dfrac y2\le x\le\sqrt{\ln3}$ is the same as $0\le x\le\sqrt{\ln3},\ 0\le y\le2x$. Hence $$\int_0^{2\sqrt{\ln3}}\int_{y/2}^{\sqrt{\ln3}}e^{x^2}\,dx\,dy=\int_0^{\sqrt{\ln3}}\int_0^{2x}e^{x^2}\,dy\,dx=\int_0^{\sqrt{\ln3}}2xe^{x^2}\,dx=\Big[e^{x^2}\Big]_0^{\sqrt{\ln3}}=e^{\ln3}-1=3-1=2.$$

(4) Since $\dfrac{1}{y^4+1}$ also has no convenient elementary antiderivative in $x$, reverse the order. The original region $0\le x\le8,\ \sqrt[3]x\le y\le2$ equals $0\le y\le2,\ 0\le x\le y^3$. Hence $$\int_0^8\int_{\sqrt[3]x}^2\frac{dy\,dx}{y^4+1}=\int_0^2\int_0^{y^3}\frac{dx\,dy}{y^4+1}=\int_0^2\frac{y^3}{y^4+1}\,dy=\Big[\tfrac14\ln(y^4+1)\Big]_0^2=\frac{\ln17}{4}.$$

(5) The region $R:0\le x\le1,\ 0\le y\le1-x$ is the same triangle when $x$ and $y$ are swapped, so by symmetry $\iint_Rx^2\,dA=\iint_Ry^2\,dA$. Since $$\iint_Rx^2\,dA=\int_0^1x^2(1-x)\,dx=\int_0^1(x^2-x^3)\,dx=\frac13-\frac14=\frac1{12},$$ we get $$\iint_R(x^2+y^2)\,dA=2\cdot\frac1{12}=\frac16.$$

(6) $R$ is the diamond enclosed by $|x|+|y|=1$. For fixed $x$, $y$ ranges over the symmetric interval $[-(1-|x|),1-|x|]$ and is an odd function of $y$, so $\iint_Ry\,dA=0$. Also $$\iint_Rx^2\,dA=\int_{-1}^1x^2\cdot2(1-|x|)\,dx=4\int_0^1x^2(1-x)\,dx=4\cdot\frac1{12}=\frac13,$$ so $$\iint_R(y-2x^2)\,dA=0-2\cdot\frac13=-\frac23.$$

Solution. (1) Integrating the inner integral in $y$ first, $$\int_{e^{-x}}^1\frac{dy}{y}=\ln1-\ln(e^{-x})=x,$$ so $$\int_1^\infty\int_{e^{-x}}^1\frac{1}{x^3y}\,dy\,dx=\int_1^\infty\frac{x}{x^3}\,dx=\int_1^\infty\frac{1}{x^2}\,dx=\Big[-\tfrac1x\Big]_1^\infty=1.$$

(2) The integrand separates in $x$ and $y$, and $\displaystyle\int_{-\infty}^\infty\frac{dx}{x^2+1}=\Big[\tan^{-1}x\Big]_{-\infty}^\infty=\pi$, so $$\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{dx\,dy}{(x^2+1)(y^2+1)}=\pi\cdot\pi=\pi^2.$$

(3) The integrand is singular at $y=1$; substituting $u=y-1$, $$\int_0^3\frac{dy}{(y-1)^{2/3}}=\int_{-1}^2\frac{du}{u^{2/3}}.$$ Here $u^{2/3}=(u^{1/3})^2$, and since the real cube root is defined for negative numbers too ($u\lt0\Rightarrow u^{1/3}\lt0$), $u^{-2/3}$ is positive for every real $u\ne0$ regardless of sign, and the integral converges near $u=0$ because the exponent $-\tfrac23\gt-1$. The antiderivative $3u^{1/3}$ (real cube root) is continuous even at $u=0$, so $$\int_{-1}^2\frac{du}{u^{2/3}}=\Big[3u^{1/3}\Big]_{-1}^2=3\cdot2^{1/3}-3\cdot(-1)=3(2^{1/3}+1).$$ Since $\displaystyle\int_0^1x^2\,dx=\frac13$, we obtain $$\int_0^1\int_0^3\frac{x^2}{(y-1)^{2/3}}\,dy\,dx=\frac13\cdot3(2^{1/3}+1)=2^{1/3}+1.$$

Solution. (1) $y=x$ and $y=x^2$ meet where $x=x^2\Leftrightarrow x(x-1)=0$, i.e., $x=0,1$. Since $x\ge x^2$ on $0\le x\le1$, the area is $$A=\int_0^1\int_{x^2}^{x}dy\,dx=\int_0^1(x-x^2)\,dx=\Big[\tfrac{x^2}2-\tfrac{x^3}3\Big]_0^1=\frac12-\frac13=\frac16.$$

(2) $y=x^2$ and $y=x+2$ meet where $x^2=x+2\Leftrightarrow(x-2)(x+1)=0$, i.e., $x=-1,2$. Since $x+2\ge x^2$ on $-1\le x\le2$, the area is $$A=\int_{-1}^2\int_{x^2}^{x+2}dy\,dx=\int_{-1}^2(x+2-x^2)\,dx=\Big[\tfrac{x^2}2+2x-\tfrac{x^3}3\Big]_{-1}^2=\frac{10}{3}-\left(-\frac76\right)=\frac92.$$

Solution. By double integral: for each $x\in[-2,2]$, the length of the $y$-interval is $(1+\sqrt{4-x^2})-(-1-\sqrt{4-x^2})=2+2\sqrt{4-x^2}$, so $$A=\int_{-2}^2\left(2+2\sqrt{4-x^2}\right)dx=\int_{-2}^22\,dx+2\int_{-2}^2\sqrt{4-x^2}\,dx=8+2\cdot2\pi=8+4\pi,$$ where $\int_{-2}^2\sqrt{4-x^2}\,dx$ is the area $2\pi$ of a semicircle of radius $2$.

Geometrically, $R$ is a rectangle of width $4$ and height $2$ ($-2\le x\le2,\ -1\le y\le1$) with a semicircle of radius $2$ attached above and below; the two semicircles together form a full circle of radius $2$ (area $4\pi$). Hence $$A=(\text{area of rectangle})+(\text{area of circle})=4\times2+4\pi=8+4\pi,$$ which agrees with the value obtained by the double integral.

Solution. The average value is $\operatorname{Avg}(f)=\dfrac{1}{A(R)}\displaystyle\iint_Rf\,dA$ with $A(R)=\pi\cdot1=\pi$. Integrating with respect to $y$ first, $$\int_0^1x\cos(xy)\,dy=\Big[\sin(xy)\Big]_{y=0}^{y=1}=\sin x,$$ so $$\iint_Rf\,dA=\int_0^\pi\sin x\,dx=\Big[-\cos x\Big]_0^\pi=1-(-1)=2.$$ Hence $$\operatorname{Avg}(f)=\frac{2}{\pi}.$$

Solution. The total number is $$N=\int_{-2}^0\int_{-5}^5\frac{10000e^y}{1+|x|/2}\,dx\,dy.$$ Since $|x|$ is even, $$\int_{-5}^5\frac{dx}{1+|x|/2}=2\int_0^5\frac{dx}{1+x/2}=2\Big[2\ln\!\left(1+\tfrac x2\right)\Big]_0^5=4\ln\frac72,$$ and $$\int_{-2}^0e^y\,dy=\Big[e^y\Big]_{-2}^0=1-e^{-2}.$$ Hence $$N=10000(1-e^{-2})\cdot4\ln\frac72=40000(1-e^{-2})\ln\frac72\approx43329\ \text{organisms}.$$

Proof. (1) By hypothesis, $m\le f(x,y)\le M$ at every point of $R$. Since $A(R)\gt0$, integrating this inequality over $R$ (by monotonicity of the integral) gives $$\iint_Rm\,dA\le\iint_Rf(x,y)\,dA\le\iint_RM\,dA,$$ that is, $$m\cdot A(R)\le\iint_Rf\,dA\le M\cdot A(R).$$ Dividing by $A(R)\gt0$ yields $$m\le\frac{1}{A(R)}\iint_Rf\,dA=\operatorname{Avg}(f)\le M.$$

(2) Suppose $R$ is open, $f\ge0$ on $R$, and $\operatorname{Avg}(f)=0$. Since $A(R)\gt0$, this forces $\iint_Rf\,dA=0$. Suppose, for contradiction, that $f(x_0,y_0)=c\gt0$ for some $(x_0,y_0)\in R$. Since $f$ is continuous at $(x_0,y_0)$, for $\dfrac c2\gt0$ there is $\delta\gt0$ such that the disk $D$ of radius $\delta$ centered at $(x_0,y_0)$ satisfies $D\subset R$ (here the openness of $R$ guarantees $D\subset R$) and $f\gt\dfrac c2$ on $D$. Writing $A(D)=\pi\delta^2\gt0$ and using $f\ge0$ on all of $R$, $$\iint_Rf\,dA=\iint_Df\,dA+\iint_{R\setminus D}f\,dA\ge\iint_Df\,dA\ge\frac c2\cdot A(D)\gt0,$$ contradicting $\iint_Rf\,dA=0$. Hence no such point exists, and $f(x,y)=0$ at every point of $R$.

Solution. The curves $r=1+\cos\theta$ (a cardioid) and $r=1$ (the unit circle) meet where $1+\cos\theta=1\Leftrightarrow\cos\theta=0\Leftrightarrow\theta=\pm\dfrac{\pi}{2}$.

The region $R$ consists of points inside $r=1+\cos\theta$ and outside $r=1$, so $1\lt r\lt1+\cos\theta$ must hold. This is possible only when $\cos\theta\gt0$, so $\theta$ ranges over $-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$, and for such $\theta$, $r$ ranges from $1$ to $1+\cos\theta$.

Since the area element in polar coordinates is $dA=r\,dr\,d\theta$, the required iterated integral is

$$\iint_Rf(r,\theta)\,dA=\int_{-\pi/2}^{\pi/2}\int_1^{1+\cos\theta}f(r,\theta)\,r\,dr\,d\theta.$$

Solution. (1) For $r^2=4\cos2\theta$ to be defined we need $\cos2\theta\ge0$, so the right-hand loop is traced for $-\dfrac{\pi}{4}\le\theta\le\dfrac{\pi}{4}$. Its area is $$A_1=\frac12\int_{-\pi/4}^{\pi/4}r^2\,d\theta=\frac12\int_{-\pi/4}^{\pi/4}4\cos2\theta\,d\theta=\Big[\sin2\theta\Big]_{-\pi/4}^{\pi/4}=1-(-1)=2.$$ The curve consists of two congruent loops symmetric about the origin, so the total area is $A=2A_1=4$.

(2) The curve $x^2+y^2=4$ is $r=2$ in polar coordinates, the line $y=1$ is $r=\csc\theta$, and the line $y=\sqrt3x$ is $\theta=\dfrac{\pi}{3}$. Since $y-\sqrt3x=r\sin\theta-\sqrt3r\cos\theta=2r\sin\!\left(\theta-\dfrac{\pi}{3}\right)$, lying above the line $y=\sqrt3x$ requires $\sin\!\left(\theta-\dfrac{\pi}{3}\right)\ge0$, i.e., $\theta\in\left[\dfrac{\pi}{3},\dfrac{4\pi}{3}\right]$. Lying above $y=1$ (with $y\ge1\gt0$) requires $r\ge\csc\theta$ and $\theta\in(0,\pi)$. Being inside the circle requires $r\le2$, i.e., $\csc\theta\le2\Leftrightarrow\sin\theta\ge\tfrac12\Leftrightarrow\theta\in\left[\dfrac{\pi}{6},\dfrac{5\pi}{6}\right]$. All three conditions together give $\theta\in\left[\dfrac{\pi}{3},\dfrac{5\pi}{6}\right]$, with $r$ ranging from $\csc\theta$ to $2$.

Hence the area is $$A=\int_{\pi/3}^{5\pi/6}\frac12\left(2^2-\csc^2\theta\right)d\theta=\Big[2\theta+\tfrac12\cot\theta\Big]_{\pi/3}^{5\pi/6},$$ and since $\cot\dfrac{5\pi}{6}=-\sqrt3$ and $\cot\dfrac{\pi}{3}=\dfrac{\sqrt3}{3}$, $$A=\left(\frac{5\pi}{3}-\frac{\sqrt3}{2}\right)-\left(\frac{2\pi}{3}+\frac{\sqrt3}{6}\right)=\pi-\frac{2\sqrt3}{3}.$$

Solution. (1) $R$ is the upper half of the unit disk ($y\ge0$), so in polar coordinates $0\le r\le1,\ 0\le\theta\le\pi$. Since $x^2+y^2=r^2$, $$\iint_Re^{x^2+y^2}\,dA=\int_0^\pi\int_0^1e^{r^2}r\,dr\,d\theta=\pi\cdot\Big[\tfrac12e^{r^2}\Big]_0^1=\pi\cdot\frac{e-1}{2}=\frac{\pi(e-1)}{2}.$$

(2) The region $x^2+y^2\le1$ is $0\le r\le1,\ 0\le\theta\le2\pi$ in polar coordinates, so the required volume is $$V=\iint(9-x^2-y^2)\,dA=\int_0^{2\pi}\int_0^1(9-r^2)r\,dr\,d\theta=2\pi\Big[\tfrac92r^2-\tfrac14r^4\Big]_0^1=2\pi\left(\frac92-\frac14\right)=2\pi\cdot\frac{17}{4}=\frac{17\pi}{2}.$$

Solution. (1) The region is the unit quarter-disk in the first quadrant, so $0\le r\le1,\ 0\le\theta\le\dfrac{\pi}{2}$ in polar coordinates. $$\int_0^1\int_0^{\sqrt{1-x^2}}(x^2+y^2)\,dy\,dx=\int_0^{\pi/2}\int_0^1r^2\cdot r\,dr\,d\theta=\frac{\pi}{2}\cdot\Big[\tfrac14r^4\Big]_0^1=\frac{\pi}{2}\cdot\frac14=\frac{\pi}{8}.$$

(2) The region is the unit quarter-disk in the third quadrant, so $0\le r\le1,\ \pi\le\theta\le\dfrac{3\pi}{2}$. $$\int_{-1}^0\int_{-\sqrt{1-x^2}}^0\frac{2}{1+\sqrt{x^2+y^2}}\,dy\,dx=\int_\pi^{3\pi/2}\int_0^1\frac{2r}{1+r}\,dr\,d\theta=\frac{\pi}{2}\int_0^1\left(2-\frac{2}{1+r}\right)dr=\frac{\pi}{2}\Big[2r-2\ln(1+r)\Big]_0^1=\frac{\pi}{2}\cdot2(1-\ln2)=\pi(1-\ln2).$$

(3) $R:1\le x^2+y^2\le e^2$ is $1\le r\le e,\ 0\le\theta\le2\pi$ in polar coordinates, and $\ln(x^2+y^2)=2\ln r$, so $$\iint_R\frac{\ln(x^2+y^2)}{\sqrt{x^2+y^2}}\,dA=\int_0^{2\pi}\int_1^e\frac{2\ln r}{r}\cdot r\,dr\,d\theta=2\pi\int_1^e2\ln r\,dr=2\pi\cdot2\Big[r\ln r-r\Big]_1^e=2\pi\cdot2\cdot1=4\pi.$$

(4) The region is the entire first quadrant, so $0\le r\lt\infty,\ 0\le\theta\le\dfrac{\pi}{2}$. $$\int_0^\infty\int_0^\infty\frac{1}{(1+x^2+y^2)^2}\,dx\,dy=\int_0^{\pi/2}\int_0^\infty\frac{r}{(1+r^2)^2}\,dr\,d\theta=\frac{\pi}{2}\cdot\left[-\frac{1}{2(1+r^2)}\right]_0^\infty=\frac{\pi}{2}\cdot\frac12=\frac{\pi}{4}.$$

Solution. The average value is $$\operatorname{Avg}(z)=\frac{1}{\pi a^2}\iint_{x^2+y^2\le a^2}\sqrt{a^2-x^2-y^2}\,dA.$$ Converting to polar coordinates, $$\iint\sqrt{a^2-r^2}\,r\,dr\,d\theta=\int_0^{2\pi}\int_0^a\sqrt{a^2-r^2}\,r\,dr\,d\theta=2\pi\left[-\frac13(a^2-r^2)^{3/2}\right]_0^a=2\pi\cdot\frac{a^3}{3}=\frac{2\pi a^3}{3},$$ so $$\operatorname{Avg}(z)=\frac{1}{\pi a^2}\cdot\frac{2\pi a^3}{3}=\frac{2a}{3}.$$

Solution. Since $u=\dfrac{2x-y}{2}=x-\dfrac{y}{2}$ and $v=\dfrac{y}{2}$, we have $y=2v$ and $x=u+v$. The Jacobian is $$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}x_u&x_v\\y_u&y_v\end{vmatrix}=\begin{vmatrix}1&1\\0&2\end{vmatrix}=2.$$

The original region is $0\le y\le4,\ \dfrac y2\le x\le1+\dfrac y2$. Since $x=\dfrac y2\Leftrightarrow u=0$ and $x=1+\dfrac y2\Leftrightarrow u=1$, and $y=2v$ so $y:0\to4$ corresponds to $v:0\to2$, the $(u,v)$-region is the rectangle $0\le u\le1,\ 0\le v\le2$.

Since the integrand is $\dfrac{2x-y}{2}=u$, $$\int_0^4\int_{y/2}^{1+y/2}\frac{2x-y}{2}\,dx\,dy=\int_0^2\int_0^1u\cdot|2|\,du\,dv=2\int_0^2\left(\int_0^1u\,du\right)dv=2\cdot\frac12\cdot2=2.$$

Solution. (1) Substituting $s=x+y,\ t=y-2x$, so $x=\dfrac{s-t}{3},\ y=\dfrac{2s+t}{3}$, the Jacobian is $$\frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac13&-\frac13\\[2pt]\frac23&\frac13\end{vmatrix}=\frac19+\frac29=\frac13.$$ On the region $0\le x,\ 0\le y,\ x+y\le1$, for fixed $s=x+y\in[0,1]$, as $x$ ranges from $0$ to $s$, $t=s-3x$ ranges from $s$ (at $x=0$) to $-2s$ (at $x=s$), so $-2s\le t\le s$.

Since the integrand is $\sqrt{x+y}(y-2x)^2=\sqrt s\,t^2$, $$\int_0^1\int_0^{1-x}\sqrt{x+y}(y-2x)^2\,dy\,dx=\int_0^1\int_{-2s}^{s}\sqrt s\,t^2\cdot\frac13\,dt\,ds=\frac13\int_0^1\sqrt s\cdot\frac{s^3-(-8s^3)}{3}\,ds=\int_0^1s^{7/2}\,ds=\Big[\tfrac29s^{9/2}\Big]_0^1=\frac29.$$

(2) Fixing $y$ and substituting $w=\sqrt{xy}$, so $x=\dfrac{w^2}{y}$, $dx=\dfrac{2w}{y}\,dw$, and $\sqrt{\dfrac{y}{x}}=\dfrac{y}{w}$. When $x=\dfrac1y$, $w=1$; when $x=y$, $w=y$. So $$\int_{1/y}^{y}\sqrt{\frac yx}\,e^{\sqrt{xy}}\,dx=\int_1^{y}\frac yw\,e^w\cdot\frac{2w}{y}\,dw=2\int_1^ye^w\,dw=2(e^y-e).$$ Hence $$\int_1^2\int_{1/y}^y\sqrt{\frac yx}\,e^{\sqrt{xy}}\,dx\,dy=\int_1^22(e^y-e)\,dy=2\Big[e^y-ey\Big]_1^2=2\big((e^2-2e)-0\big)=2e^2-4e=2e(e-2).$$

(3) Following the hint, substitute $x=u^2-v^2,\ y=2uv$ $(u,v\ge0)$. Then $x^2+y^2=(u^2+v^2)^2$, so $\sqrt{x^2+y^2}=u^2+v^2$, and the Jacobian is $$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}2u&-2v\\2v&2u\end{vmatrix}=4u^2+4v^2=4(u^2+v^2).$$

Substituting into the boundary curve $y=2\sqrt{1-x}$, i.e., $y^2=4(1-x)$, gives $$4u^2v^2+4(u^2-v^2)=4\ \Longleftrightarrow\ u^2(v^2+1)=v^2+1\ \Longleftrightarrow\ u^2=1\ \Longleftrightarrow\ u=1,$$ so this entire curve corresponds to $u=1$. Also $y=0$ corresponds to $v=0$, and $x=0$ corresponds to $u=v$. Hence the original region $\{0\le x\le1,\ 0\le y\le2\sqrt{1-x}\}$ corresponds, in the $(u,v)$-plane, to the triangle $0\le v\le u\le1$ with vertices $(0,0),(1,0),(1,1)$.

Since the product of the integrand and the Jacobian is $(u^2+v^2)\cdot4(u^2+v^2)=4(u^2+v^2)^2$, $$\int_0^1\int_0^{2\sqrt{1-x}}\sqrt{x^2+y^2}\,dy\,dx=\int_0^1\int_0^u4(u^2+v^2)^2\,dv\,du.$$ The inner integral is $$\int_0^u4(u^4+2u^2v^2+v^4)\,dv=4\left(u^5+\frac23u^5+\frac15u^5\right)=4\cdot\frac{28}{15}u^5=\frac{112}{15}u^5,$$ so $$\int_0^1\frac{112}{15}u^5\,du=\frac{112}{15}\cdot\frac16=\frac{56}{45}.$$