answer:Given: - Sphere radius, r = 0.055 m (converting 11 cm to meters) - Electric potential on the sphere, V = 12,000 V - Coulomb's constant, k = 8.99 × 10^9 N·m^2/C^2 - Permittivity of free space, ε₀ = 8.854 × 10^-12 F/m The charge on the sphere, q, can be related to the electric potential using: V = k * (q / r) The electric field (E) near the surface of the sphere is given by: E = k * (q / r^2) Simplifying using the potential: E = V / r E = 12,000 V / 0.055 m E ≈ 218,181.82 N/C The energy density (U) in the electric field is: U = (1/2) * ε₀ * E^2 U = (1/2) * (8.854 × 10^-12) * (218,181.82)^2 U ≈ 0.211 J/m^3 Therefore, the energy density in the electric field near the surface of the sphere is approximately 0.211 J/m^3.

answer:The function f(x) is not one-to-one, so it does not have a unique inverse function. Therefore, we cannot directly evaluate the integral in terms of f^{-1}(x). However, we can use the fact that f(x) is a quartic polynomial to estimate the value of the integral. Since f(x) is positive for large x, we have f^{-1}(x) approx x^{1/4} for large x. Therefore, intlimits_{2a}^{8a} frac{1}{(f^{-1}(x))^2 + (f^{-1}(x))^4} dx approx intlimits_{2a}^{8a} frac{1}{x^{1/2} + x} dx = left[ 2ln(x^{1/2} + x) right]_{2a}^{8a} = 2lnleft(frac{1+2sqrt{2a}}{1+sqrt{2a}}right) As a to infty, this expression approaches 2ln(2). Therefore, we have limlimits_{a to infty} intlimits_{2a}^{8a} frac{1}{(f^{-1}(x))^2 + (f^{-1}(x))^4} dx = 2ln(2) = ln(2^2) = ln(4) Therefore, n = boxed{4}.

answer:The Jacobian matrix of the vector-valued function mathbf{r}(x, y, z) is given by: J(mathbf{r}) = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} Evaluating the partial derivatives, we get: J(mathbf{r}) = begin{bmatrix} -4(y - x)^3 & 4(y - x)^3 & 0 0 & 1 & 0 frac{3}{2(z - x)^{5/2}} & 0 & -frac{3}{2(z - x)^{5/2}} end{bmatrix}

answer:Let f_n(x)=dfrac{x^n(ax^2+ax+1)}{e^x}, with xin[0,1]. If x is fixed in [0,1], then we have lim_{nto infty}f_n(x)=begin{cases}0&&text{if }xin[0,1)dfrac{2a+1}e&&text{if }x=1 end{cases} so we have pointwise convergence of f_n in [0,1] to f(x):=lim_{ntoinfty}f_n. Now, let's check for uniform convergence. We have |f_n(x)-f(x)|=left|frac{x^n(ax^2+ax+1)}{e^x}-frac{2a+1}{e}right| =frac{|x^n(ax^2+ax+1)-(2a+1)|}{e^x} =frac{|x^n(ax^2+ax+1-2a-1)|}{e^x} =frac{|x^n(ax^2+(a-2)x)|}{|e^x|} =frac{|x^n||ax^2+(a-2)x|}{|e^x|} =frac{x^n|ax^2+(a-2)x|}{e^x} =frac{x^n|x||ax+a-2|}{e^x} =frac{x^{n+1}|ax+a-2|}{e^x} Since 0le xle 1, we have 0le x^{n+1}le 1. Also, since a is a constant, |ax+a-2|le |a|+|a-2|=2|a|+2. Therefore, |f_n(x)-f(x)|le frac{2|a|+2}{e^x} Now, let's find lim_{ntoinfty}sup_{xin[0,1]}|f_n(x)-f(x)|. We have lim_{ntoinfty}sup_{xin[0,1]}|f_n(x)-f(x)|le lim_{ntoinfty}frac{2|a|+2}{e^x} =frac{2|a|+2}{e^1} =frac{2|a|+2}{e} Since frac{2|a|+2}{e} is a constant, we have lim_{ntoinfty}sup_{xin[0,1]}|f_n(x)-f(x)|=0 Therefore, f_n converges uniformly to f on [0,1]. Finally, we can evaluate the limit of the integral: lim_{ntoinfty}int_0^1f_n(x)dx=int_0^1lim_{ntoinfty}f_n(x)dx=int_0^1f(x)dx =int_0^1frac{2a+1}{e}dx =frac{2a+1}{e}int_0^1dx =frac{2a+1}{e}cdot 1 =frac{2a+1}{e} Therefore, the answer is boxed{textbf{d) }frac{2a+1}{e}}.