Calcular as seguintes integrais reiteradas:
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d \varphi \int_{0}^{3 \cos \varphi} r^{2} \operatorname{sen}^{2} \varphi d r.

Passo 1

O enunciado pede para calcular a integral iterada
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\!\!\Bigl[\int_{0}^{3\cos\varphi}r^{2}\sin^{2}\varphi\,dr\Bigr]d\varphi.
O conteúdo é integração iterada em coordenadas polares (ou em variáveis r,\varphi). Vamos primeiro integrar em r e depois em \varphi, usando simetria para simplificar.

Passo 2

Calculemos a integral interna em r:
\int_{0}^{3\cos\varphi}r^{2}\sin^{2}\varphi\,dr =\sin^{2}\varphi\;\int_{0}^{3\cos\varphi}r^{2}\,dr =\sin^{2}\varphi\;\biggl[\frac{r^{3}}{3}\biggr]_{0}^{3\cos\varphi} =\sin^{2}\varphi\;\frac{(3\cos\varphi)^{3}}{3} =9\cos^{3}\varphi\,\sin^{2}\varphi.

Passo 3

Substituindo na integral exterior:
I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}9\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi =9\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi.
Como \cos^{3}\varphi\,\sin^{2}\varphi é função par, temos
I=9\cdot2\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi =18\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi.

Passo 4

Escrevemos \sin^{2}\varphi=1-\cos^{2}\varphi e separamos:
\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi =\int_{0}^{\frac{\pi}{2}}(\cos^{3}\varphi-\cos^{5}\varphi)\,d\varphi =\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,d\varphi -\int_{0}^{\frac{\pi}{2}}\cos^{5}\varphi\,d\varphi.

Passo 5

Usamos a fórmula de redução \displaystyle\int_{0}^{\frac{\pi}{2}}\cos^{n}\varphi\,d\varphi=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}\cos^{n-2}\varphi\,d\varphi. Assim,
\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,d\varphi =\frac{2}{3}\int_{0}^{\frac{\pi}{2}}\cos^{1}\varphi\,d\varphi =\frac{2}{3}\Bigl[\sin\varphi\Bigr]_{0}^{\frac{\pi}{2}} =\frac{2}{3},
\int_{0}^{\frac{\pi}{2}}\cos^{5}\varphi\,d\varphi =\frac{4}{5}\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,d\varphi =\frac{4}{5}\cdot\frac{2}{3} =\frac{8}{15}.

Passo 6

Substituindo esses valores:
\int_{0}^{\frac{\pi}{2}}\cos^{3}\varphi\,\sin^{2}\varphi\,d\varphi =\frac{2}{3}-\frac{8}{15} =\frac{10-8}{15} =\frac{2}{15},
logo
I=18\cdot\frac{2}{15} =\frac{36}{15} =\frac{12}{5}.

Resposta

\displaystyle \frac{12}{5}

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