Integración: ¿qué debo hacer ahora?

Question

$$\iint_\Omega (x^2+3y^3)\mathrm dx \mathrm dy $$ $$\Omega: 0\le x^2+y^2\le 1$$ Lo integré dos veces y luego obtuve esto

$$Cy+\frac{x^2y^2}{4}+xy+c $$

¿Qué debo hacer ahora?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \iint_{x^{2} + y^{2} < 1}\pars{x^{2} + y^{3}}\,\dd x\,\dd y & = {1 \over 2}\iint_{x^{2} + y^{2} < 1}r^{2}\,\dd x\,\dd y = {1 \over 2}\int_{0}^{2\pi}\int_{0}^{1}r^{2}\,r\,\dd r\,\dd\theta = \pi\int_{0}^{1}r^{3}\,\dd r\,\dd\theta \\[5mm] & = \bbx{\pi \over 4} \end{align}

---

$$\mbox{Note that}\quad \left\{\substack{\ds{\iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y\,\,\, \stackrel{y\ \mapsto\ -y}{=}\,\,\, -\iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y \implies \iint_{x^{2} + y^{2} < 1}y^{3}\,\dd x\,\dd y = \color{#f00}{0}} \\[1cm] \ds{\iint_{x^{2} + y^{2} < 1}x^{2}\,\dd x\,\dd y\,\,\, \stackrel{x\ \leftrightarrow\ y}{=}\,\,\, \iint_{x^{2} + y^{2} < 1}y^{2}\,\dd x\,\dd y} \\[3mm] \ds{\implies \iint_{x^{2} + y^{2} < 1}x^{2}\,\dd x\,\dd y = {1 \over 2}\iint_{x^{2} + y^{2} < 1} \pars{x^{2} + y^{2}}\,\dd x\,\dd y = {1 \over 2}\iint_{x^{2} + y^{2} < 1}r^{2} \,\dd x\,\dd y}}\right. $$

Answer 2

Alternativamente:

$$\int \int_{\Omega} x^2+3y^3 dxdy=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2+3y^3dydx=\int_{-1}^1 (x^2y+\frac{3}{4}y^4)\big|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx=\int_{-1}^12x^2\sqrt{1-x^2}dx \stackrel{x=\sin t}=\int_{-\pi/2}^{\pi/2}2\sin^2t\cdot\cos^2tdt=\int_{-\pi/2}^{\pi/2}\frac12\sin^2 2tdt=\int_{-\pi/2}^{\pi/2}\frac{1}{4}(1-\cos4t)dt=\frac{\pi}{4}.$$

Answer 3

\begin{align} &\int \int_{\Omega} x^2+3y^3 dxdy \\&= \int\int_{\Omega} x^2dxdy \\&=4 \int_0^{\frac{\pi}2}\int_0^{1}r^3\cos^2(\theta) drd\theta \\&=4 \int_0^{\frac{\pi}2}\cos^2(\theta)d\theta\int_0^{1}r^3 dr \\ &= \int_0^{\frac{\pi}2}\cos^2(\theta) d\theta \\ &= \int_0^{\frac{\pi}2} \frac{\cos(2\theta)+1}{2}d\theta \\ &= \left[ \frac{\sin(2\theta)}4+\frac{\theta}2\right]_0^{\frac{\pi}2} \\ &=\frac{\pi}4 \end{align}

Observación:

Podrías revisar tus pasos de integración en tu trabajo, me resulta contraintuitivo que obtengas términos de bajo orden.