Respuesta :
Answer:
Step-by-step explanation:
The equation of the sphere, centered a the origin is given by [tex]x^2+y^2+z^2 = 64[/tex]. Then, when z=4, we get
[tex]x^2+y^2= 64-16 = 48 [/tex].
This equation corresponds to a circle of radius [tex]4\sqrt[]{3}[/tex] in the x-y plane
c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from [tex]-4\sqrt[]{3}[/tex] up to [tex]4\sqrt[]{3}[/tex]. This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from [tex]-\sqrt[]{48-x^2}[/tex] and [tex]\sqrt[]{48-x^2}[/tex], this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to [tex]\sqrt[]{64-x^2-y^2}[/tex]
Then, the triple integral that gives us the volume of D in cartesian coordinates is
[tex]\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx[/tex].
b) Recall that the cylindrical coordinates are given by [tex]x=r\cos \theta, y = r\sin \theta,z = z[/tex], where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that [tex]x^2+y^2 = r^2[/tex]. WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since [tex]\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta[/tex] goes from 0 to [tex]2\pi[/tex]. Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume
[tex]\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta[/tex]. Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).
a) For the spherical coordinates, recall that [tex]z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta [/tex]. where [tex]\phi[/tex] is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4[tex]\sqrt[]{3}[/tex] if we take the positive square root of 48. So, let us calculate the angle between the vector[tex]a=(0,4\sqrt[]{3},4)[/tex] and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula
[tex]\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}[/tex]
Therefore, over the circle, [tex]\phi = \frac{\pi}{3}[/tex]. Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since [tex]z = \rho \cos \phi[/tex], then over the plane we have that [tex]\rho = \frac{4}{\cos \phi}[/tex] Then, the following is the desired integral
[tex]\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta[/tex] where the new factor is the jacobian for the spherical coordinates.
d ) Let us use the integral in cylindrical coordinates
[tex]\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr[/tex]
Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution [tex]u = 64-r^2[/tex] then [tex]\frac{-du}{2} = r dr[/tex], then
[tex]=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}[/tex]
