Question

Find the surface area of the part of the sphere

x^2+y^2+z^2=9

lying above the ellipse

x^2/4 + y^2/9 =1

Answer 1

$$S=\int\int_D\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}dxdy$$

$$x^2+y^2+z^2=9$$

$$z^2=9-x^2-y^2$$

$$2z(\partial z/\partial x)=-2x$$

$$\partial z/\partial x=-x/z$$

$$2z(\partial z/\partial y)=-2y$$

$$\partial z/\partial y=-y/z$$

$$1+(\partial z/\partial x)^2+(\partial z/\partial y)^2=1+\dfrac{x^2}{z^2}+\dfrac{y^2}{z^2}$$

$$=\dfrac{z^2+x^2+y^2}{z^2}=\dfrac{9}{z^2}$$

$$z^2=9-x^2-y^2, z\geq0$$

$$S=\int\int_D\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}dxdy$$

$$=\int\int_D\dfrac{3}{\sqrt{9-x^2-y^2}}dxdy$$

$$\dfrac{x^2}{4}+\dfrac{y^2}{9}\leq1$$

$$x=2r\cos\theta, y=3r\sin\theta$$

$$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix} 2\cos\theta & -2r\sin\theta \\ 3\sin\theta & 3r\cos\theta \end{vmatrix}=6r$$

$$S=\int\int_G\dfrac{6r}{\sqrt{9-4r^2\cos^2\theta-9r^2\sin^2\theta}}drd\theta$$

$$=\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{1}\dfrac{6r}{\sqrt{9-4r^2-5r^2\sin^2\theta}}drd\theta$$

$$=6\displaystyle\int_{0}^{2\pi}\bigg(\dfrac{3-\sqrt{5}\sin\theta}{4+5\cos^2\theta}\bigg)d\theta$$

$$=6\pi$$

$S=6\pi$ square units