La integral

Question

<blockquote> <p>Calcular la integral <span class="math-container">$$ \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3) dy dx .$ $</span></p> </blockquote> <p>Mi intento: que <span class="math-container">$ x^2y^3 $</span> es una función impar con respecto a los <span class="math-container">$ y $</span>, tan <span class="math-container">\begin{align*} \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+x^2y^3)\, dy dx &= \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}( x^2 )\, dy dx \\ &=\int_{0}^{1} 2x^2\sqrt{1-x^2}\, dx\\ &= \int_{0}^{\frac{\pi}{2}}2\sin^2\theta\cos^2\theta \,d\theta\\ &= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin^22\theta \,d\theta\\ &= \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-\cos 4\theta) \,d\theta\\ &= \frac{\pi}{8}.\end{align*}</span></p> <p>¿Estoy correcto? ¿Creo que ha un poco demasiado complicado que lo que debería ser?</p>

Answer 1

Está bien.

Tal vez sería computar usando coordenadas polares:<span class="math-container">$$\int_{-\frac\pi2}^\frac\pi2\int0^1r^3\cos^2(\theta)+r^6\cos^2(\theta)\sin^3(\theta)\,\mathrm dr\,\mathrm d\theta.$$Using the same argument as yours, this is just<span class="math-container">$$\int{-\frac\pi2}^\frac\pi2\int_0^1r^3\cos^2(\theta)\,\mathrm dr\,\mathrm d\theta,$$</span>which is equal to<span class="math-container">$% $ $\left(\int0^1r^3\,\mathrm dr\right)\left(\int{-\frac\pi2}^\frac\pi2\cos^2(\theta)\,\mathrm d\theta\right)=\frac\pi8.$</span></span>