Surface Area

[JJ007]

Find the area of the portion of the sphere \(\displaystyle x^2+y^2+z^2=9\) that lies inside the cylinder\(\displaystyle x^2+y^2=3y\)

\(\displaystyle S=\int_R\int\sqrt{1+[f_x(x,y)]^2+[f_y(x,y)]^2}dA\)

Not sure what to do.

Thanks

 

[BigGlenntheHeavy]

\(\displaystyle Quite \ a \ protracted \ one, \ however \ I'll \ take \ a \ stab \ at \ it.\)

\(\displaystyle Find \ the \ area \ of \ the \ portion \ of \ the \ sphere \ x^2+y^2+z^2 \ = \ 9 \ that \ lies \ inside \ the\)

\(\displaystyle \ cylinder \ x^2+y^2 \ = \ 3y. \ Note, \ z \ = \ f(x,y)\)

\(\displaystyle Ergo, \ S \ = \ \int_R\int\sqrt{1+[f_x(x,y)]^2+[f_y(x,y)]^2} dA \ = \ \int_R\int\sqrt{1+(z_x)^2+(z_y)^2} dA\)

\(\displaystyle now \ z^2 \ = \ 9-x^2+y^2, \ z \ = \ \pm\sqrt{9-x^2-y^2}\)

\(\displaystyle z_x \ = \ \pm\frac{1}{2}(9-x^2-y^2)^{-1/2}(-2x) \ = \ \pm\frac{-x}{(9-x^2-y^2)^{1/2}} \ = \ \bigg(\frac{-x}{z}\bigg).\)

\(\displaystyle z_y \ = \ \pm\frac{1}{2}(9-x^2-y^2)^{-1/2}(-2y) \ = \ \pm\frac{-y}{(9-x^2-y^2)^{1/2}} \ = \ \bigg(\frac{-y}{z}\bigg).\)

\(\displaystyle z_x^2+z_y^2+1 \ = \ \frac{x^2}{z^2}+\frac{y^2}{z^2}+1 \ = \ \frac{x^2+y^2+z^2}{z^2} \ = \ \frac{9}{9-x^2-y^2}\)

\(\displaystyle and \ x^2+y^2 \ = \ 3y \ \implies \ r^2 \ = \ 3rsin(\theta), \ \implies \ r \ = \ 3sin(\theta).\)

\(\displaystyle Hence, \ S \ = \ (2)(2)\int_{0}^{\pi/2}\int_{0}^{3sin(\theta)}\frac{3r}{\sqrt{9-r^2}} drd\theta \ = \ 18(\pi-2) \ sq. \ units\)

\(\displaystyle Here \ is \ its \ graph.\)

[attachment=0:3dl6s1ta]iii.jpg[/attachment:3dl6s1ta]

 

[JJ007]

\(\displaystyle Quite \ a \ protracted \ one, \ however \ I'll \ take \ a \ stab \ at \ it.\)

\(\displaystyle Find \ the \ area \ of \ the \ portion \ of \ the \ sphere \ x^2+y^2+z^2 \ = \ 9 \ that \ lies \ inside \ the\)

\(\displaystyle \ cylinder \ x^2+y^2 \ = \ 3y. \ Note, \ z \ = \ f(x,y)\)

\(\displaystyle Ergo, \ S \ = \ \int_R\int\sqrt{1+[f_x(x,y)]^2+[f_y(x,y)]^2} dA \ = \ \int_R\int\sqrt{1+(z_x)^2+(z_y)^2} dA\)

\(\displaystyle now \ z^2 \ = \ 9-x^2+y^2, \ z \ = \ \pm\sqrt{9-x^2-y^2}\)

\(\displaystyle z_x \ = \ \pm\frac{1}{2}(9-x^2-y^2)^{-1/2}(-2x) \ = \ \pm\frac{-x}{(9-x^2-y^2)^{1/2}} \ = \ \bigg(\frac{-x}{z}\bigg).\)

\(\displaystyle z_y \ = \ \pm\frac{1}{2}(9-x^2-y^2)^{-1/2}(-2y) \ = \ \pm\frac{-y}{(9-x^2-y^2)^{1/2}} \ = \ \bigg(\frac{-y}{z}\bigg).\)

\(\displaystyle z_x^2+z_y^2+1 \ = \ \frac{x^2}{z^2}+\frac{y^2}{z^2}+1 \ = \ \frac{x^2+y^2+z^2}{z^2} \ = \ \frac{9}{9-x^2-y^2}\)

\(\displaystyle and \ x^2+y^2 \ = \ 3y \ \implies \ r^2 \ = \ 3rsin(\theta), \ \implies \ r \ = \ 3sin(\theta).\)

\(\displaystyle Hence, \ S \ = \ (2)(2)\int_{0}^{\pi/2}\int_{0}^{3sin(\theta)}\frac{3r}{\sqrt{9-r^2}} drd\theta\)

\(\displaystyle I \ think \ I'm \ right \ up \ to \ here, \ can \ someone \ finished \ it \ off?\)

No need. I think that's good enough.
Thanks a lot!