Multiple Dimensions - Calculate The Mass Of The Sun

We can assume the element to be concentric hollow spheres, having the same center as the sun. Let mass of the sun be $M$ (in $\text{kg}$ ), and the radius be $\text{R}$ .( $1 \text{R} = 6.90 \times 10^{10} \text{cm}$ ).

For any general sphere with radius $x$ , with thickness $dx$ , its mass $dM$ can be written as density times volume, i.e

$dM = D \times 4 \pi x^2 dx$

$D$ is in $\text{g / cm}^3$ , However, $x$ is not in $\text{cm}$ , $x$ is in $\text{R}$ .

Let's convert $D$ to $\text{kg/R}^3$ (by multiplying $(6.90 \times 10^{10})^3$ and dividing by $1000$ ) to make calculations correct, and get final answer in $\text{kg}$ .

$dM = (6.90 \times 10^{10})^3 \times \frac{1}{1000} \times (519x^4 - 1630x^3 + 1844x^2 - 889x + 155) \times 4 \pi x^2 dx$

$M = \int dM = 4\pi\times(328.5 \times 10^{30}) \times \frac{1}{1000}\int (519x^6 - 1630x^5 + 1844x^4 - 889x^3 + 155x^2) dx$

Since $x$ goes from $0$ to $1$ , the limits on the integral will be $0$ and $1$

$M = 4\pi \times (328.5 \times 10^{27}) \int_0^1 (519x^6 - 1630x^5 + 1844x^4 - 889x^3 + 155x^2) dx$

$M =4\pi \times (3.29 \times 10^{29})\times\left(\frac{519}{7} - \frac{1630}{6} + \frac{1844}{5} - \frac{889}{4} + \frac{155}{3}\right)$

$M = 4\pi \times 3.29 \times 10^{29} \times \frac{97}{140}$

$M = \boxed{2.86 \times 10^{30}\text{kg}}$