Respuesta :
Since it is in the first octant, if we let [tex]\theta[/tex] be the angle formed in the xy plane, then its limits are simply [tex]0 \leq \theta \leq \frac{\pi}{2}[/tex].
The radius is a constant of [tex]\sqrt{5}[/tex] so we integrate from [tex0 \leq \rho \leq \sqrt 5[/tex]
The yz/xz plane doesn't go the full 90 degrees. Instead, it goes to the [tex]z=2[/tex] plane which means that it forms a triangle of hypotenuse [tex]\sqrt{5}[/tex] and an opposite leg of [tex]2[/tex]. This produces an angle of [tex]\phi = 1.107[/tex] so our limits for [tex]0 \leq \phi \leq 1.107[/tex]
We are just integrating a constant of 1, then we get:
[tex]\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1.107}\int_{0}^{\sqrt{5}} d\rho d\phi d\theta[/tex]
The radius is a constant of [tex]\sqrt{5}[/tex] so we integrate from [tex0 \leq \rho \leq \sqrt 5[/tex]
The yz/xz plane doesn't go the full 90 degrees. Instead, it goes to the [tex]z=2[/tex] plane which means that it forms a triangle of hypotenuse [tex]\sqrt{5}[/tex] and an opposite leg of [tex]2[/tex]. This produces an angle of [tex]\phi = 1.107[/tex] so our limits for [tex]0 \leq \phi \leq 1.107[/tex]
We are just integrating a constant of 1, then we get:
[tex]\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1.107}\int_{0}^{\sqrt{5}} d\rho d\phi d\theta[/tex]
A triple integral including limits of integration that gives the volume of the cap of the solid sphere x2+y2+z2≤5 cut off by the plane z=2 is [tex]\rm \int\limits^{\pi /4}_0\int\limits^{\pi /2}_0\int\limits^{\sqrt{2}} _{sec\theta} \rho^{2} sin\theta \:d\rho d\theta d\delta[/tex]
What is triple integration?
The triple integral of a function f(x,y,z) over a rectangular box B is defined as
[tex]f ( x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)\,\Delta x \Delta y \Delta z = \iiint_B f(x,y,z) \,dV\] if this limit exists.[/tex]
Since it is in the first octant, if we let the angle formed in the XY plane, then its limits are simple.
[tex]x^{2} +y^{2} +z^{2} =2[/tex]
[tex]\rho^{2} = 2[/tex]
[tex]\rho = \sqrt{2}[/tex]
z = 2
[tex]\rho cos \theta =2\\\rho = 2sec \theta[/tex]
[tex]2sec \theta = \rho = 2\\[/tex]
[tex]cos\theta = 1/\sqrt{2}\\\theta = \pi /4[/tex]
The triple integral
[tex]\rm \int\limits^{\pi /4}_0\int\limits^{\pi /2}_0\int\limits^{\sqrt{2}} _{sec\theta} \rho^{2} sin\theta \:d\rho d\theta d\delta[/tex]
The yz/xz plane doesn't go the full 90 degrees. Instead, it goes to the plane which means that it forms a triangle of the hypotenuse and an opposite leg.
Learn more about triple integration;
https://brainly.com/question/2289273
