91Ó°ÊÓ

Firstly, let's find the volume of the cone using triple integral in spherical coordinates. Converting \(z^2 = x^2 + y^2\) into cylindrical coordinates, we get: $$\rho^2 = z^2$$ Here, \(0 \leq \rho \leq z\), \(0 \leq \theta \leq 2\pi\), and \(0\leq z\leq 2\). Therefore, the volume of the cone, \(V\), in terms of cylindrical coordinates is given by: $$V = \int\limits_0^2 \int\limits_0^{2\pi} \int\limits_0^z \rho\, \text{d}\rho\, \text{d}\theta\, \text{d}z$$ Now let's evaluate the integral: $$ V = \int\limits_0^2\int\limits_0^{2\pi}\frac{z^3}{2}\text{d}\theta \text{d}z $$ $$ V = \int\limits_0^2 z^3 \left[\int\limits_0^{2\pi}\text{d}\theta\right] \text{d}z $$ $$ V = \int_0^2 z^3 (2\pi) \text{d}z $$ $$ V= 2\pi\left[\frac{z^4}{4}\right]_0^2 $$ $$ V = 2\pi\left(4\right)=8\pi $$ The volume of the cone is \(V = 8\pi\).

Using cylindrical coordinates for the triple integral, we get: $$\int\limits_0^2 \int\limits_0^{2\pi} \int\limits_0^z (100 - 25z) \rho\, \text{d}\rho\, \text{d}\theta\, \text{d}z$$ Let's evaluate the integral: $$ \int\limits_0^2\int\limits_0^{2\pi}\frac{(100 - 25z) z^3}{2}\text{d}\theta \text{d}z $$ $$ \int\limits_0^2(100 - 25z) z^3\left[\int\limits_0^{2\pi}\text{d}\theta\right] \text{d}z $$ $$ \int\limits_0^2 (100 - 25z) z^3 (2\pi) \text{d}z $$ $$ 2\pi\left[\int\limits_0^2 (100z^3 - 25z^4) \text{d}z\right] $$ $$ 2\pi\left[\frac{100z^4}{4} - \frac{25z^5}{5}\right]_0^2 $$ The result of the integral is \(800\pi\).

Now divide the result of the integral with the volume of the cone: $$ \text{Average value} = \frac{800\pi}{8\pi} $$ $$ \text{Average value} = 100 $$ The average value of the temperature function \(T(x, y, z) = 100 - 25z\) on the cone \(z^2 = x^2 + y^2\) for \(0 \leq z \leq 2\) is 100.