Na elipsoidzie
: 22 sie 2018, o 00:54
Na elipsoidzie \(\displaystyle{ E=\left\{ \left( x,y,z\right) \in \RR^3:x^2+y^2+4z^2=4 \right\}}\) rozważamy zbiór \(\displaystyle{ A=\left\{\left( x,y,z\right) \in \RR^3:y>x,y>0,z>0 \right\}}\). Obliczyć \(\displaystyle{ \int_{A}^{}xyzd\sigma_2}\).
Proszę o sprawdzenie poniższego rozwiązania:
\(\displaystyle{ z= \sqrt{1-1/4x^2-1/4y^2}}\)
\(\displaystyle{ z'_x= -\frac{x}{ \sqrt{16-4x^2-4y^2} }}\)
\(\displaystyle{ z'_y= -\frac{y}{ \sqrt{16-4x^2-4y^2} }}\)
\(\displaystyle{ \int_{A}^{}xyzd\sigma_2= \int_{D}^{}xy \sqrt{1-1/4x^2-1/4y^2} \cdot \sqrt{1+\frac{x^2}{ 16-4x^2-4y^2 }+\frac{x^2}{ 16-4x^2-4y^2 }}dxdy=}\)
\(\displaystyle{ \int_{D}^{}xy/4 \sqrt{16-3x^2-3y^2}dxdy=}\)
Podstawiam \(\displaystyle{ x=r\cos \phi,y=r\sin \phi, J=r}\)
\(\displaystyle{ = \int_{\pi/4}^{\pi} \int_{0}^{2}1/4r^2\sin \phi\cos \phi \sqrt{16-3r^2}rdrd\phi=}\)
\(\displaystyle{ = \int_{\pi/4}^{\pi}1/4\sin \phi\cos\phi(-16/27(16-3r^2)^{3/2}+1/45(16-3r^2)^{5/2}d\phi}\) w granicy od zera do dwóch. To się równia:
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(-128/27+32/45+1024/27-1024/45)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(896/27-992/45)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(1504/135)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}376/135\sin \phi\cos \phi d\phi=}\)
\(\displaystyle{ 188/135\int_{\pi/4}^{\pi}\sin 2\phi d\phi=-94/135}\)
Czy tak jest dobrze?
Proszę o sprawdzenie poniższego rozwiązania:
\(\displaystyle{ z= \sqrt{1-1/4x^2-1/4y^2}}\)
\(\displaystyle{ z'_x= -\frac{x}{ \sqrt{16-4x^2-4y^2} }}\)
\(\displaystyle{ z'_y= -\frac{y}{ \sqrt{16-4x^2-4y^2} }}\)
\(\displaystyle{ \int_{A}^{}xyzd\sigma_2= \int_{D}^{}xy \sqrt{1-1/4x^2-1/4y^2} \cdot \sqrt{1+\frac{x^2}{ 16-4x^2-4y^2 }+\frac{x^2}{ 16-4x^2-4y^2 }}dxdy=}\)
\(\displaystyle{ \int_{D}^{}xy/4 \sqrt{16-3x^2-3y^2}dxdy=}\)
Podstawiam \(\displaystyle{ x=r\cos \phi,y=r\sin \phi, J=r}\)
\(\displaystyle{ = \int_{\pi/4}^{\pi} \int_{0}^{2}1/4r^2\sin \phi\cos \phi \sqrt{16-3r^2}rdrd\phi=}\)
\(\displaystyle{ = \int_{\pi/4}^{\pi}1/4\sin \phi\cos\phi(-16/27(16-3r^2)^{3/2}+1/45(16-3r^2)^{5/2}d\phi}\) w granicy od zera do dwóch. To się równia:
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(-128/27+32/45+1024/27-1024/45)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(896/27-992/45)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}1/4\sin \phi\cos \phi(1504/135)d\phi=}\)
\(\displaystyle{ \int_{\pi/4}^{\pi}376/135\sin \phi\cos \phi d\phi=}\)
\(\displaystyle{ 188/135\int_{\pi/4}^{\pi}\sin 2\phi d\phi=-94/135}\)
Czy tak jest dobrze?