Question

Resolvendo a integral:

Obtemos:

Answer 1

$\displaystyle\int_0^3\int_0^{\sqrt{9-x^2}}dy\,dx\\\\\\ =\int_0^3 y\big|_0^{\sqrt{9-x^2}}\,dx\\\\\\ =\int_0^3 (\sqrt{9-x^2}-0^2)\,dx\\\\\\ =\int_0^3 \sqrt{9-x^2}\,dx$

Substituição trigonométrica:

$x=3\,\mathrm{sen\,}t~~\Rightarrow~~\left\{ \begin{array}{l} dx=3\cos t\,dt\\\\ -\frac{\pi}{2}\le t\le \frac{\pi}{2} \end{array}\right.\\\\\\ \sqrt{9-x^2}=\sqrt{9-9\,\mathrm{sen^2\,}t}\\\\ =\sqrt{9\,(1-\mathrm{sen^2\,}t)}\\\\ =\sqrt{9\cos^2 t}\\\\ =3\left|\cos t\right|~~~~~~\big(\text{mas }-\frac{\pi}{2}\le t\le \frac{\pi}{2}\big)\\\\ =3\cos t$

Mudando os extremos de integração:

$\text{Quando }x=0~~\Rightarrow~~t=0\\\\ \text{Quando }x=3~~\Rightarrow~~t=\dfrac{\pi}{2}$

Substituindo a integral fica

$=\displaystyle\int_0^{\pi/2} 3\cos t\cdot 3\cos t\,dt\\\\\\ =\int_0^{\pi/2} 9\cos^2 t\,dt\\\\\\ =\int_0^{\pi/2} 9\left(\dfrac{1}{2}+\dfrac{1}{2}\cos(2t) \right )dt\\\\\\ =9\cdot \left.\left(\dfrac{t}{2}+\dfrac{1}{4}\,\mathrm{sen}(2t) \right )\right|_0^{\pi/2}\\\\\\ =9\cdot \left[\left(\dfrac{\left(\frac{\pi}{2} \right )}{2}+\dfrac{1}{4}\,\mathrm{sen}(\pi)\right)-\left(\dfrac{t}{2}+\dfrac{1}{4}\,\mathrm{sen\,}(0)\right) \right ]\\\\\\ =\dfrac{9\pi}{4}$

$\therefore~~\boxed{\begin{array}{c}\displaystyle\int_0^3\int_0^{\sqrt{9-x^2}}dy\,dx=\dfrac{9\pi}{4} \end{array}}$