Volumen de un cono en coordenadas cilíndricas

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Hola ana paola!

Si r= a  altura h, el resultado será:

$$\begin{align}&V= \frac{\pi} 3 a^2 h\\&\\&Ecuación \ del \ cono:\\&\\&x^2+y^2-k^2z^2=0\\&En \ cilíndricas:\\&\\&r^2-k^2z^2=0\ ==> z= \frac r k\\&\theta \in [0,2 \pi]\\&\\&r \in [0,a]\\&\\&z \in [ \frac r k, h]\\& \\&Cuando\ r=a ==> z=h \ \Rightarrow k= \frac r z= \frac a h \Rightarrow  \frac r k= \frac{ rh } a\\&\\&|Jacobiano|=r\\&\\&V= \int_0^{2 \pi}  \int_{r=0}^{r=a}  \int_{z=  \frac{rh} a}^h\ \ r \ dz\ dr \ d \theta=\\&\\&\int_0^{2 \pi}  \int_{r=0}^{r=a}  \Bigg[\int_{z=  \frac{rh} a}^h\ \  \ dz \Bigg] \ r \  dr \ d \theta=\\&\\&\\&\int_0^{2 \pi}  \int_{r=0}^{r=a}  \Bigg [z \Bigg]_{z=  \frac{rh} a}^h\ \ r \ \ dr \ d \theta=\\&\\&\\&\\&\int_0^{2 \pi}  \int_{r=0}^{r=a}   \Bigg(h- \frac {rh} a \Bigg)\ \ r \ \ dr \ d \theta=\\&\\&\int_0^{2 \pi}  \int_{r=0}^{r=a}   \Bigg(hr- \frac {r^2h} a \Bigg)\ \ \ \ dr \ d \theta=\\&\\& \\&h\int_0^{2 \pi} \Bigg[ \int_{r=0}^{r=a} \Big( r- \frac{r^2} a\big)dr \Bigg] d \theta=\\&\\&h \int_0^{2 \pi} \  \Bigg[ \frac{r^2} 2- \frac{r^3}{3a} \Bigg]_0^a\ d \theta= h \int_0^{2 \pi} \ \Bigg( \frac {a^2} 2- \frac{a^3}{3a} \Bigg) d \theta=\\&\\&h \int_0^{2 \pi} \ \Big( \frac{a^2}2- \frac {a^2}3 \Big) d \theta=\\&\\&h \int_0^{2 \pi} \  \frac 1 6 a^2 d \theta= \frac 1 6 a^2 h  \int_0^{2 \pi} \ d \theta= \frac 1 6 a^2h \ \ \theta  \Bigg|_0^{2 \pi}\\&\\&= \frac 1 6 a^2 h  2 \pi= \frac{\pi} 3 a^2 h\\&\\&\end{align}$$

Saludos y recuerda votar

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