Triple integral (tplquad) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a triple integral using tplquad

What is the output of this code that calculates the triple integral of f(x,y,z) = x + y + z over the cube where x, y, z range from 0 to 1?

SciPy

from scipy.integrate import tplquad

func = lambda z, y, x: x + y + z
result, error = tplquad(func, 0, 1, lambda x: 0, lambda x: 1, lambda x, y: 0, lambda x, y: 1)
print(round(result, 3))

A 2.0

B 3.0

C 0.5

D 1.5

💡 Hint

Remember the integral of x, y, or z from 0 to 1 is 0.5 each, and the integral sums over all three variables.

The integral of x from 0 to 1 is 0.5, same for y and z. Summing these gives 1.5 as the total integral over the unit cube.

❓ data_output

intermediate

Number of items in the result tuple from tplquad

After running tplquad, how many items are in the returned result tuple?

SciPy

from scipy.integrate import tplquad

func = lambda z, y, x: x*y*z
result = tplquad(func, 0, 1, lambda x: 0, lambda x: 1, lambda x, y: 0, lambda x, y: 1)
print(len(result))

A 2

B 1

C 3

D 4

💡 Hint

tplquad returns the integral value and an estimate of the error.

tplquad returns a tuple with two items: the integral result and the estimated error.

🔧 Debug

advanced

Identify the error in this tplquad usage

What error does this code raise when trying to compute the triple integral of f(x,y,z) = x*y*z with incorrect bounds?

SciPy

from scipy.integrate import tplquad

func = lambda z, y, x: x*y*z
result = tplquad(func, 0, 1, lambda x: 1, lambda x: 0, lambda x, y: 0, lambda x, y: 1)
print(result[0])

A TypeError: 'int' object is not callable

B SyntaxError: invalid syntax

C ValueError: Integration limits are not in ascending order

D No error, outputs a float

💡 Hint

Check the order of the lower and upper bounds in the lambda functions.

The inner integral bounds must be in ascending order. Here, the lower bound lambda returns 1 and upper bound returns 0, causing a ValueError.

🧠 Conceptual

advanced

Understanding the order of variables in tplquad

In the function passed to tplquad, what is the order of variables for the integrand function?

A y, z, x

B z, y, x

C x, y, z

D x, z, y

💡 Hint

Check the scipy documentation for tplquad variable order in the integrand.

The integrand function for tplquad receives variables in the order z, y, x, where x is the outermost variable.

🚀 Application

expert

Calculate the volume of a tetrahedron using tplquad

Use tplquad to calculate the volume of the tetrahedron bounded by x, y, z ≥ 0 and x + y + z ≤ 1. What is the integral result?

SciPy

from scipy.integrate import tplquad

func = lambda z, y, x: 1
result, error = tplquad(func, 0, 1, lambda x: 0, lambda x: 1 - x, lambda x, y: 0, lambda x, y: 1 - x - y)
print(round(result, 3))

A 0.167

B 0.333

C 0.500

D 1.000

💡 Hint

The volume of this tetrahedron is 1/6, which is approximately 0.167.

The volume of the tetrahedron defined by x,y,z≥0 and x+y+z≤1 is 1/6. The integral of 1 over this region equals this volume.