Answer to Question #96326 in Calculus for Rachel

Question

Set up a triple integral that represents the volume of a cylinder of radius R and height h in (a) Cartesian coordinates, (b) cylindrical coordinates and (c) spherical coordinates.

Answer 1

By definition, the volume of a cylinder is given by the formula: $V = \oiiint_C \limits dV$ . Consider that cylinder is placed along the z-axis with the center of base in the origin of coordinate system.

(a) Cartesian coordinates.

$V = \iiint_C \limits dxdydz$ .

The region of integration: $-R \le x\le R, -\sqrt{R^2-x^2} \le y\le \sqrt{R^2-x^2}, 0\le z\le h.\\$

Thus:

$V = \int_0^h \limits dz \int_{-R}^R \limits dx \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\limits dy = \pi R^2 h$

(b)Cylindrical coordinates.

Coordinate transformation is given by: $x = \rho \cos{\varphi}, y = \rho \sin{\varphi}, z = z$ .

The region of integration: $0 \le \rho\le R,\: 0\le \varphi\le2\pi,\: 0\le z\le h\\$

The Jacobian (see https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant):

$J = \rho \Rightarrow dxdydz = \rho\cdot d\rho d\varphi dz$.

Thus:

$V = \iiint_C \limits \rho\cdot d\rho d\varphi dz = \int_0^h \limits dz \int_{0}^{2\pi} \limits d\varphi \int_0^R \limits \rho d\rho = \pi R^2 h$

(c) Spherical coordinates.

Coordinate transformation is given by: $x = \rho\sin{\theta} \cos{\varphi},\: y = \rho\sin{\theta} \sin{\varphi}, z = \rho\cos{\theta}$.

The region of integration. The equation of the limited cylindrical surface in spherical coordinates is: $\rho \sin{\theta}=R,\: \arctan{\dfrac{R}{h}} \le \theta \le \dfrac{\pi}{2}$. The equation of the top base of cylinder is: $\rho \cos{\theta}=h,\: 0 \le \theta \le \arctan{\dfrac{R}{h}}$ .

And, due to z-symmetry, $0\le \varphi\le2\pi$.

The Jacobian of transformation: $J = \rho^2 \sin{\theta} \Rightarrow dxdydz = \rho^2 \sin{\theta}\cdot d\rho d\varphi d\theta$.

Thus:

$V = \iiint_C \limits \rho^2 \sin{\theta}\cdot d\varphi d\theta d\rho = \int_{0}^{2\pi} \limits d\varphi \left[ \int_0^{\arctan{\frac{R}{h}}} \limits \sin{\theta} d\theta \int_0^{\frac{h}{\cos{\theta}}} \limits \rho^2 d\rho + \int_{\arctan{\frac{R}{h}}}^{\frac{\pi}{2}} \limits \sin{\theta} d\theta \int_0^{\frac{R}{\sin{\theta}}} \limits \rho^2 d\rho \right] = \pi R^2 h$

P.S. Integrating over the cylinder in spherical coordinates is not convenient.