Answer to Question #192513 in Calculus for Nikhil Rawat

Question #192513

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


1
Expert's answer
2021-05-18T18:11:50-0400
S=D1+(z/x)2+(z/y)2dxdyS=\int\int_D\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}dxdy

x2+y2+z2=9x^2+y^2+z^2=9

z2=9x2y2z^2=9-x^2-y^2

2z(z/x)=2x2z(\partial z/\partial x)=-2x

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

2z(z/y)=2y2z(\partial z/\partial y)=-2y

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

1+(z/x)2+(z/y)2=1+x2z2+y2z21+(\partial z/\partial x)^2+(\partial z/\partial y)^2=1+\dfrac{x^2}{z^2}+\dfrac{y^2}{z^2}

=z2+x2+y2z2=9z2=\dfrac{z^2+x^2+y^2}{z^2}=\dfrac{9}{z^2}

z2=9x2y2,z0z^2=9-x^2-y^2, z\geq0

S=D1+(z/x)2+(z/y)2dxdyS=\int\int_D\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}dxdy

=D39x2y2dxdy=\int\int_D\dfrac{3}{\sqrt{9-x^2-y^2}}dxdy

x24+y291\dfrac{x^2}{4}+\dfrac{y^2}{9}\leq1

x=2rcosθ,y=3rsinθx=2r\cos\theta, y=3r\sin\theta

(x,y)(r,θ)=2cosθ2rsinθ3sinθ3rcosθ=6r\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=G6r94r2cos2θ9r2sin2θdrdθS=\int\int_G\dfrac{6r}{\sqrt{9-4r^2\cos^2\theta-9r^2\sin^2\theta}}drd\theta


=02π016r94r25r2sin2θdrdθ=\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{1}\dfrac{6r}{\sqrt{9-4r^2-5r^2\sin^2\theta}}drd\theta

=602π(35sinθ4+5cos2θ)dθ=6\displaystyle\int_{0}^{2\pi}\bigg(\dfrac{3-\sqrt{5}\sin\theta}{4+5\cos^2\theta}\bigg)d\theta

=6π=6\pi

S=6πS=6\pi square units



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