Respuesta :
[tex]x^2+y^2+z^2=100z\iff x^2+y^2+(z-50)^2=50^2[/tex]
The surface of interest can be parameterized by
[tex]\mathbf r(u,v)=\begin{cases}x(u,v)=50\cos u\sin v\\y(u,v)=50\sin u\sin v\\z(u,v)=50(\cos v+1)\end{cases}[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\arccos\dfrac{49}{50}[/tex].
The area of the surface [tex]S[/tex] is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dA=\iint_T\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv[/tex]
We have
[tex]\|\mathbf r_u\times\mathbf r_v\|=50^2\sin v[/tex]
so the area is given by the double integral
[tex]\displaystyle50^2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\arccos(49/50)}\sin v\,\mathrm dv\,\mathrm du=100\pi[/tex]
The surface of interest can be parameterized by
[tex]\mathbf r(u,v)=\begin{cases}x(u,v)=50\cos u\sin v\\y(u,v)=50\sin u\sin v\\z(u,v)=50(\cos v+1)\end{cases}[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\arccos\dfrac{49}{50}[/tex].
The area of the surface [tex]S[/tex] is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dA=\iint_T\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv[/tex]
We have
[tex]\|\mathbf r_u\times\mathbf r_v\|=50^2\sin v[/tex]
so the area is given by the double integral
[tex]\displaystyle50^2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\arccos(49/50)}\sin v\,\mathrm dv\,\mathrm du=100\pi[/tex]
The area of the part of the sphere [tex]x^2+y^2+z^2=100z[/tex] that lies inside the paraboloid [tex]z=x^2+y^2[/tex] is 100[tex]\pi[/tex] and this can be determined by doing the integration.
Given :
- The equation of sphere -- [tex]x^2+y^2+z^2=100z[/tex]
- Equation of paraboloid -- [tex]z=x^2+y^2[/tex]
First, parameterize S as given below:
[tex]\rm r(u,v) = \left\{ \begin{matrix}x(u,v)\; =\;50\;cosu\;sinv & & \\ y(u,v)\;= \;50\;sinu\;sinv & & \\ z(u,v)\;= \;50\;(cosv+1) & & \end{matrix}\right.[/tex]
In the above expression the value of [tex]\rm 0\leq u\leq 2\pi[/tex] and [tex]\rm 0\leq v\leq cos^{-1}\frac{49}{50}[/tex]
Now, evaluate the surface integral as given below:
[tex]\rm \int\int_SdA = \int\int_T||r_u\times r_v|| \; du\;dv[/tex] --- (1)
Now, the value of [tex]\rm ||r_u\times r_v||[/tex] is given by:
[tex]\rm ||r_u\times r_v|| = 50^2sin\;v[/tex]
Now, substitute the value of [tex]\rm ||r_u\times r_v||[/tex] in the expression (1).
[tex]\rm \int\int_SdA = 50^2\int^{2\pi}_0\int^{cos^{-1}\frac{49}{50}}_0 sinv \; dv\;du[/tex]
Simplify the above integral in order to determine the area of the part of the sphere.
[tex]\rm \int\int_SdA = 100\pi[/tex]
For more information, refer to the link given below:
https://brainly.com/question/14502499
