6.11
Calculate the area of R and \(\displaystyle\iint_{R} \left(\frac{2y^{4}}{x^{2}}+ \frac{3y^{2}}{x}\right) e^{xy}dA\) for \(R\) the region bounded by the curves \(y^{2} = 3x, y^{2} = 2x, xy= 2\) and \(xy = 1\).
We want a change of variables \(G(u,v)\) where \(G: R_{0} \to R\) bijective and \(R_0 = [a,b] \times [c,d]\).
To find \(G\) we can find the inverse of \(G\), say \(F\). Then \(F(x,y)=(u,v)\) where \(F: R \to R_{0}\)
We want \(v(x,y)\) such that \(F\) carries the hyperbolas \(xy=1\) to \(c=v\) and \(xy=2\) to \(v=d\).
Define \(v(x,y) = xy\), \(1 \le v\le 2\).
We want \(F\) to carry the parabola \(y^2=2x\) to the line \(u=a\) and \(y^2=3x\) to \(u=b\).
Set \(u(x,y) = \frac{y^{2}}{x}\), \(2 \le u \le 3\).
We get \(F(x,y) = (\frac{y^2}{x}, xy)\). To compute the integrals (area(R)). We need \(|J_G|\)
Since \(G = F^{-1}\), then \(|J_G| = \frac{1}{|J_F|}\), then \(J_F = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} -\frac{y^2}{x^2} & \frac{2y}{x} \\ y & x \end{vmatrix}\) \(= -\frac{y^2}{x} - \frac{2y^2}{x} = -\frac{3y^2}{x}\) \(= -3u\)
So \(|J_G| = |\frac{1}{-3u}| = \frac{1}{3u}\)
\(\text{area}(R) = \iint_{R} dA = \int_{1}^{2} \int_{2}^{3} \frac{1}{3u} du dv\) \(= \int_{1}^{2} \frac{1}{3} (\ln(3) - \ln(2)) dv\) \(= \int_{1}^{2} \frac{1}{3} \ln(\frac{3}{2}) dv = \frac{1}{3} \ln(\frac{3}{2})\)
\(\iint_{R} (\frac{2y^4}{x^2} + \frac{3y^2}{x}) e^{xy} dA = \int_{1}^{2} \int_{2}^{3} (2u^2 + 3u) e^v \frac{1}{3u} du dv\) \(= \int_{1}^{2} \int_{2}^{3} (\frac{2}{3}u + 1) e^v du dv\) \(= \int_{1}^{2} e^v [\frac{u^2}{3} + u]_{2}^{3} dv\)
\(= \int_{1}^{2} \frac{8}{3} e^v dv\) \(= \frac{8}{3} [e^v]_{1}^{2} = \frac{8}{3} (e^2 - e)\)
Compute \(\iint_D xye^{-x^2-y^2} dA\) where \(D\) is the first quadrant of \(\mathbb{R}^2\).
\(f(x,y)=xye^{-x^2-y^2}\) is non-negative on \(D\).
Consider \(\{D_n\}, D_n \subset D \xrightarrow{\text{bounded}}\) such that \(f\) is integrable on \(D_n\) and \(D_n \xrightarrow{n \to \infty} D\).
\(f\) is integrable on \(D\) if \(\lim_{n \to \infty} \iint_{D_n} f dA\) exists.
Consider \(D_n = \{(x,y) \in \mathbb{R}^2: x \in [0,n], y \in [0,n]\}\) where \(D_n \xrightarrow{n} D\)
\(f\) is continuous on \(D_n\), so it's integrable on \(D_n\).
\(\iint_{D_{n}}xye^{-x^2-y^2}dA=\int_0^{n}\int_0^{n}(xe^{-x^2})(ye^{-y^2})dxdy\) since \((ye^{-y^2})\int_0^{n}xe^{-x^2}dx=(ye^{-y^2})\left(-\frac12e^{-x^2}\right)_0^{n}\)
Then \(\int_0^{n}ye^{-y^2}\left(-\frac12e^{-n^2}+\frac12\right)dy=\left(-\frac12e^{-n^2}+\frac12\right)\left(-\frac12e^{-y^2}\right)_0^{n}\)
Then \(\lim_{n\to\infty}\left(-\frac{1}{2}e^{-n^2}+\frac{1}{2}\right)\left(-\frac{1}{2} e^{-n^2}+\frac{1}{2}\right)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)
Show that \(\iint_D \frac{1}{(x^2+y^2)^5+1} dA\) is convergent, where \(D\) is the first quadrant of \(\mathbb{R}^2\).
Consider \(D_n = \{(x,y) \in \mathbb{R}^2 : x \ge 0, y \ge 0, x^2+y^2 \le n^2\}\)
\(D_n \underset{n \to \infty}{\longrightarrow} D\), then \(f(x,y) = \frac{1}{(x^2+y^2)^5+1}\) is continuous and so integrable on \(D_n\).
So we compute \(\lim_{n \to \infty} \iint_{D_n} f dA\).
Let's use polar coordinates. For \(D_n\), \(0 \le \theta \le \frac{\pi}{2}\), \(0 \le r \le n\).
So \(\iint_{D_n} f dA = \int_0^{\pi/2} \int_0^n \frac{r}{(r^2)^5+1} dr d\theta = \frac{\pi}{2} \int_0^n \frac{r}{r^{10}+1} dr\).
We look for \(\lim_{n \to \infty} \frac{\pi}{2} \int_0^n \frac{r}{r^{10}+1} dr = L\) or \(\int_0^\infty \frac{r}{r^{10}+1} dr\) to converge.
\(\int_0^\infty \frac{r}{r^{10}+1} dr = \int_0^1 \frac{r}{r^{10}+1} dr + \int_1^\infty \frac{r}{r^{10}+1} dr\).
Since \(\int_0^1 \frac{r}{r^{10}+1} dr\) exists, all we have to do is to check that \(\int_1^\infty \frac{r}{r^{10}+1} dr\) converges.
Comparison test \(0 \leq f(r) \leq g(r)\) and \(\int_1^\infty g(r)dr\) converges, then \(\int_1^\infty f(r)dr\) converges.
Since \(\frac{r}{r^{10}+1} \leq \frac{r}{r^{10}}\). Now, \(\int_1^\infty \frac{r}{r^{10}}dr = \int_1^\infty \frac{1}{r^9}dr\) (P-test), \(q>1\) so the integral converges
Then by comparison test, we have that \(\int_1^\infty \frac{r}{r^{10}+1}dr\) converges.
Thus \(\iint_D \frac{1}{(x^2+y^2)^5+1}dA\) converges.
Show that \(\iint_{D}\frac{y-x}{(x+y+1)((x^{2}+y^{2})^{5}+1)}dA\) is convergent, where \(D\) is the first quadrant of \(\mathbb{R}^2\).
Note that the function takes negative values on \(D\).
So, we use the fact that if \(\iint_D |f| dA < \infty \implies \iint_D f dA < \infty\) and \(0 \le g(x,y) \le h(x,y)\), then \(\iint_D h(x,y) dA\) converges \(\implies \iint_D g dA\) converges.
We have \(|\frac{Y-X}{(x+y+1)((x^2+y^2)^5+1)}| \le \frac{|Y|+|X|}{(x+y+1)((x^2+y^2)^5+1)}\) \(\le \frac{Y+X+1}{(x+y+1)((x^2+y^2)^5+1)} = \frac{1}{(x^2+y^2)^5+1}\)
We've seen that \(\int_D \frac{1}{(x^2+y^2)^5+1} dA < \infty\). Then \(\iint_D |\frac{Y-X}{(x+y+1)((x^2+y^2)^5+1)}| dA < \infty\)
\(\implies\) the original integral converges.
Consider \(\iint_D \frac{x^2+1}{\sqrt[3]{y-1}} dA\) where D is the rectangle \(0 \le x \le 1, 0 \le y \le 2\).
Let \(D = D_1 \cup D_2\) since the function is not bounded in \(y=1\)
If we check that \(\iint_{D_1} \frac{x^2+1}{\sqrt[3]{y-1}} dA \text{ converges}\) for \(i=1,2\), then \(\iint_D \frac{x^2+1}{\sqrt[3]{y-1}}\) converges.
For \(D_2\), the function is non-negative and is discontinuous on the boundary of \(D_2\).
\(\iint_{D_2} \frac{x^2+1}{\sqrt[3]{y-1}} dA = \int_0^1 \int_1^2 \frac{x^2+1}{\sqrt[3]{y-1}} dy dx \text{ and you can check this integral exists.}\)
The same for \(D_1\)
Consider \(f(x,y) = \begin{cases} \frac{XY(X^2-Y^2)}{(X^2+Y^2)^3} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y)=(0,0) \end{cases}\) and \(\iint_D f dA\), for \(D = \{(x,y) \in \mathbb{R}^2: 0 \leq x \leq 2, 0 \leq y \leq 1\}\).
The function is not continuous at \((0,0)\). Now \(\int_0^2 \int_0^1 \frac{XY(X^2-Y^2)}{(X^2+Y^2)^3} dy dx\)
For the inner\(\int_0^1 \frac{xy(x^2-y^2)}{(x^2+y^2)^3} dy\) \(= \int_{x^2}^{x^2+1} \frac{x(2x^2-u)}{2u^3} du\) where \(u = x^2+y^2\) and \(du = 2y dy\)
\(= \int_{x^2}^{x^2+1} \left(\frac{x^3}{u^3} - \frac{x}{2u^2}\right) du\) \(= \frac{x}{2(x^2+1)^2}\)
So, \(\int_0^2 \int_0^1 \frac{xy(x^2-y^2)}{(x^2+y^2)^3} dy dx = \int_0^2 \frac{x}{2(x^2+1)^2} dx\) \(= \left. \frac{1}{4(x^2+1)} \right|_0^2 = \frac{1}{5}\)
But, if you compute, you can check that \(\int_0^1 \int_0^2 \frac{xy(x^2-y^2)}{(x^2+y^2)^3} dxdy = -\frac{1}{20} \neq \frac{1}{5}\)
The result is true for non-negative function, not general