Volume do parabolóide \( z = 25 - x^2 - y^2 \) acima do plano \(xy\)

Solução

Usaremos coordenadas polares:

\[ x = r\cos\theta,\quad y = r\sin\theta,\quad x^2 + y^2 = r^2 \]

A função se torna:

\[ z = 25 - r^2 \]

O domínio é:

\[ 0 \leq r \leq 5,\quad 0 \leq \theta \leq 2\pi \]

Cálculo do Volume

\[ V = \iint_D (25 - x^2 - y^2)\, dA = \int_0^{2\pi} \int_0^5 (25 - r^2) \cdot r\, dr\, d\theta \]

Integral interna:

\[ \int_0^5 (25r - r^3)\, dr = \left[ \frac{25}{2}r^2 - \frac{1}{4}r^4 \right]_0^5 = \frac{25}{2}(25) - \frac{1}{4}(625) = \frac{625}{2} - \frac{625}{4} = \frac{625}{4} \]

Integral externa:

\[ V = \int_0^{2\pi} \frac{625}{4}\, d\theta = \frac{625}{4} \cdot 2\pi = \frac{1250\pi}{4} = \frac{625\pi}{2} \]

Resposta final: \( \boxed{V = \dfrac{625\pi}{2}} \)