Evaluate The Following Integral In Cylindrical Coordinates

Introduction

Evaluating a triple integral by switching to cylindrical coordinates often turns a daunting calculation into a straightforward exercise. The method is especially powerful when the region of integration possesses rotational symmetry around the (z)-axis or when the integrand itself contains the combination (x^{2}+y^{2}). In this article we will evaluate the integral

[ \iiint_{E} f(x,y,z),dV ]

by converting the Cartesian description of the domain (E) into cylindrical coordinates ((r,\theta ,z)). We will walk through the geometric intuition, the algebraic conversion, and a complete worked‑out example that exceeds 900 words, ensuring that every step is crystal clear for students, engineers, and anyone who needs a reliable reference for future problems.

Why Cylindrical Coordinates?

Feature	Cartesian ((x,y,z))	Cylindrical ((r,\theta ,z))
Symmetry handling	Awkward for rotational symmetry	Natural for circles, cylinders, cones
Volume element	(dx,dy,dz)	(r,dr,d\theta ,dz)
Common integrands	Polynomials in (x,y)	Functions of (r) or (r^{2}=x^{2}+y^{2})
Visualisation	Rectangular boxes	Stacked “tubes” and “disks”

When the region (E) can be described by simple inequalities in (r) and (\theta), the Jacobian determinant of the transformation contributes a factor of (r) to the differential volume, simplifying the integral dramatically.

The Cylindrical Coordinate Transformation

The transformation from Cartesian to cylindrical coordinates is defined by

[ \begin{cases} x = r\cos\theta,\[4pt] y = r\sin\theta,\[4pt] z = z, \end{cases} \qquad \begin{cases} r = \sqrt{x^{2}+y^{2}},\[4pt] \theta = \tan^{-1}!\left(\dfrac{y}{x}\right),\[4pt] z = z. \end{cases} ]

The Jacobian determinant (\displaystyle\left|\frac{\partial(x,y,z)}{\partial(r,\theta ,z)}\right|) equals (r). Consequently

[ dV = dx,dy,dz = r,dr,d\theta ,dz. ]

Any integrand (f(x,y,z)) can be rewritten as (f(r\cos\theta , r\sin\theta , z)). The new limits for (r,\theta ,z) are obtained by projecting the original region onto the (r\theta)-plane and the (z)-axis.

Step‑by‑Step Procedure

Describe the region (E) in Cartesian form.
Identify the bounding surfaces (planes, spheres, cylinders, etc.).
Sketch the region (even a rough sketch helps to see the symmetry).
Translate the bounding equations into cylindrical language.
- Replace (x^{2}+y^{2}) by (r^{2}).
- Replace (x) and (y) by (r\cos\theta) and (r\sin\theta).
Determine the limits for (r), (\theta), and (z).
- (\theta) usually runs from (0) to (2\pi) (full rotation) or a sub‑interval if the region is a sector.
- (r) runs from an inner radius to an outer radius, possibly depending on (z) or (\theta).
- (z) limits are taken from the lowest to the highest surface after solving for (z) in terms of (r) (or vice‑versa).
Rewrite the integrand in terms of (r,\theta ,z).
Insert the Jacobian factor (r). The integral becomes

[ \iiint_{E} f(x,y,z),dV = \int_{\theta_{\min}}^{\theta_{\max}} \int_{r_{\min}(\theta)}^{r_{\max}(\theta)} \int_{z_{\min}(r,\theta)}^{z_{\max}(r,\theta)} f(r\cos\theta , r\sin\theta , z); r , dz,dr,d\theta . ]
Integrate in the order that yields the simplest antiderivatives—often (dz) first, then (dr), then (d\theta).
Check units and symmetry to ensure no sign errors or missed factors.

Worked Example

Problem: Evaluate

[ I = \iiint_{E} (x^{2}+y^{2}),dV, ]

where (E) is the solid bounded below by the plane (z=0) and above by the paraboloid (z = 4 - x^{2} - y^{2}).

1. Identify the region

The paraboloid intersects the plane (z=0) when

[ 0 = 4 - x^{2} - y^{2};\Longrightarrow; x^{2}+y^{2}=4. ]

Thus the projection onto the (xy)-plane is the disk (x^{2}+y^{2}\le 4) (radius (2)). The solid is a paraboloid cap opening downward.

2. Convert to cylindrical coordinates

(x^{2}+y^{2}=r^{2}).
The upper surface becomes (z = 4 - r^{2}).
The lower surface is (z = 0).

The radial coordinate runs from (0) to the intersection radius (2). The angular coordinate covers a full revolution: (0\le\theta\le2\pi).

Want to learn more? We recommend with another person or thing and why should you learn a language for further reading.

3. Rewrite the integrand

(x^{2}+y^{2}=r^{2}). The Jacobian contributes an extra factor (r). Hence

[ I = \int_{0}^{2\pi}!Which means \int_{0}^{2}! \int_{0}^{,4-r^{2}} r^{2}; r , dz,dr,d\theta = \int_{0}^{2\pi}!Plus, \int_{0}^{2}! \int_{0}^{,4-r^{2}} r^{3}, dz,dr,d\theta .

4. Perform the innermost integration (with respect to (z))

[ \int_{0}^{,4-r^{2}} r^{3},dz = r^{3}\bigl[,z,\bigr]_{0}^{4-r^{2}} = r^{3},(4-r^{2}) = 4r^{3} - r^{5}. ]

5. Integrate with respect to (r)

[ \int_{0}^{2} \bigl(4r^{3} - r^{5}\bigr),dr = 4\int_{0}^{2} r^{3},dr - \int_{0}^{2} r^{5},dr = 4\left[\frac{r^{4}}{4}\right]{0}^{2} - \left[\frac{r^{6}}{6}\right]{0}^{2} = \left[r^{4}\right]{0}^{2} - \frac{1}{6}, \left[r^{6}\right]{0}^{2}. ]

Compute the powers:

[ r^{4}\big|{2}=2^{4}=16,\qquad r^{6}\big|{2}=2^{6}=64. ]

Thus

[ \int_{0}^{2} (4r^{3} - r^{5}),dr = 16 - \frac{64}{6} = 16 - \frac{32}{3} = \frac{48}{3} - \frac{32}{3} = \frac{16}{3}. ]

6. Integrate with respect to (\theta)

[ I = \int_{0}^{2\pi} \frac{16}{3},d\theta = \frac{16}{3},\bigl[,\theta,\bigr]_{0}^{2\pi} = \frac{16}{3},(2\pi) = \boxed{\dfrac{32\pi}{3}}. ]

Result: The value of the integral is (\displaystyle \frac{32\pi}{3}).

Notice how the cylindrical transformation reduced a three‑dimensional Cartesian integral into a product of simple one‑dimensional integrals, thanks to the symmetry of the region.

Scientific Explanation of the Jacobian

The factor (r) that appears in the volume element (r,dr,d\theta,dz) has a clear geometric meaning. But imagine a tiny rectangular box in Cartesian space with side lengths (dx,dy,dz). Even so, when we rotate this box around the (z)-axis, the sides parallel to the (x)‑ and (y)-axes sweep out an annular sector whose area is approximately the product of the radial thickness (dr) and the arc length (r,d\theta). Multiplying by the vertical thickness (dz) yields the volume (r,dr,d\theta,dz). Put another way, the Jacobian compensates for the fact that circles farther from the origin have larger circumferences.

Frequently Asked Questions

Q1: When should I prefer cylindrical coordinates over spherical coordinates?

A: Use cylindrical coordinates when the region is bounded by cylinders, planes parallel to the (z)-axis, or surfaces expressed as (z = f(r)). Spherical coordinates are better suited for spheres and cones that emanate from the origin.

Q2: What if the region is a sector, not a full disk?

A: Adjust the (\theta)-limits to the appropriate angular interval, e.g., (\theta\in[\alpha,\beta]) where (\beta-\alpha) equals the sector’s angle. All other steps remain identical.

Q3: How do I handle integrands that contain (z) multiplied by (x^{2}+y^{2})?

A: Replace (x^{2}+y^{2}) with (r^{2}) and keep (z) unchanged. The integrand becomes (z,r^{2}) and the volume element adds another factor (r), giving (z,r^{3}) inside the integral.

Q4: Can the order of integration be changed arbitrarily?

A: In principle yes, but the limits may become more complicated. Choose the order that yields the simplest inner limits—often (dz) first because the (z)-bounds are usually expressed directly in terms of (r).

Q5: What if the Jacobian is zero at (r=0)? Does that cause problems?

A: The Jacobian vanishes only at the axis, a set of measure zero. It does not affect the value of the integral because the contribution from a single line (or point) is zero. The details matter here.

Common Pitfalls and How to Avoid Them

Pitfall	Description	Remedy
Forgetting the Jacobian	Omitting the factor (r) leads to an answer that is too small by a factor of the average radius.	Always write (dV = r,dr,d\theta,dz) before starting the integration.
Mis‑identifying limits for (r)	Using the outer radius for both inner and outer limits when the region has a hole (e.Day to day, g. Here's the thing — , a cylindrical shell).	Sketch the cross‑section; determine inner radius (r_{\text{in}}(z)) and outer radius (r_{\text{out}}(z)) separately. Which means
Mixing up (\theta) range	Using (0) to (\pi) for a full solid of revolution.	Remember full revolution is (0) to (2\pi); only restrict (\theta) if the solid is a sector.
Incorrect conversion of surfaces	Substituting (x^{2}+y^{2}=r^{2}) but forgetting to replace isolated (x) or (y).	Replace every occurrence of (x) and (y) with (r\cos\theta) and (r\sin\theta) respectively.
Ignoring sign of (z) limits	Swapping upper and lower bounds when the surface is given as (z = g(r)) with (g) decreasing.	Write the limits explicitly: (z_{\min}= \text{lower surface},; z_{\max}= \text{upper surface}).

Conclusion

Switching to cylindrical coordinates is a powerful technique for evaluating triple integrals when the geometry of the region or the form of the integrand suggests rotational symmetry. By following a systematic procedure—identifying the region, converting bounding equations, determining limits, inserting the Jacobian, and integrating in a convenient order—one can transform a seemingly complex Cartesian integral into a manageable calculation, as demonstrated in the worked example that yielded (\displaystyle \frac{32\pi}{3}).

Remember that the key to success lies in a clear visualisation of the solid, careful translation of inequalities, and never forgetting the (r) factor that accounts for the expanding circumference of circles as you move away from the axis. Master these steps, and cylindrical coordinates will become a natural, almost automatic tool in your mathematical toolbox, ready to simplify problems ranging from physics (electric fields of cylindrical charge distributions) to engineering (heat flow in pipes) and beyond.