Find the volume of a sphere using spherical coordinates Find the volume of the sphere $x^2 +y^2+z^2 \le 4$ that are between planes $z= 1/3$ and $z=-1/4$ using spherical coordinates

My attempt:

I know that $\theta$ would be from  $0\le\theta\le 2\pi $
If $z_i=\frac{-1}{3}+\frac{i}{n}\cdot(\frac{1}{4}-\frac{-1}{3})=\frac{-1}{3}+\frac{7i}{12n},i\in \left \{ 1,2,...,n \right \}$, then $R_i=(4-z_i^2)^{0.5}$, so:

$V=\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{1}{n}\cdot \pi R_i^2 =\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{1}{n}\cdot \pi (4-(\frac{-1}{3}+\frac{7i}{12n})^2)$.

Let $f(x)=\pi(4-x^2)$, then $V=\frac{12}{7}\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{7}{12n}\cdot f(\frac{-1}{3}+\frac{7i}{12n})=\frac{12}{7}\int_{\frac{-1}{3}}^{\frac{1}{4}}f(x).dx=\frac{12}{7}\int_{\frac{-1}{3}}^{\frac{1}{4}}\pi(4-x^2).dx=\frac{12}{7}(4\pi x-\frac{\pi}{3}x^3)_{\frac{-1}{3}}^{\frac{1}{4}}=\frac{12}{7}(\pi-\frac{\pi}{192}+\frac{4\pi}{3}-\frac{\pi}{81})$
If $z_i=\frac{-1}{3}+\frac{i}{n}\cdot(\frac{1}{4}-\frac{-1}{3})=\frac{-1}{3}+\frac{7i}{12n},i\in \left \{ 1,2,...,n \right \}$, then $R_i=(4-z_i^2)^{0.5}$, so:

$V=\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{1}{n}\cdot \pi R_i^2 =\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{1}{n}\cdot \pi (4-(\frac{-1}{3}+\frac{7i}{12n})^2)$.

Let $f(x)=\pi(4-x^2)$, then $V=\frac{12}{7}\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{7}{12n}\cdot f(\frac{-1}{3}+\frac{7i}{12n})=\frac{12}{7}\int_{\frac{-1}{3}}^{\frac{1}{4}}f(x).dx=\frac{12}{7}\int_{\frac{-1}{3}}^{\frac{1}{4}}\pi(4-x^2).dx=\frac{12}{7}(4\pi x-\frac{\pi}{3}x^3)_{\frac{-1}{3}}^{\frac{1}{4}}=\frac{12}{7}(\pi-\frac{\pi}{192}+\frac{4\pi}{3}-\frac{\pi}{81})$