# Evaluate triple integral

#### TheMercury79

Problem Statement
I have evaluated this integral and got 16. But when I do it in the Mathematica program, the answer comes out as 32. I'm wondering if there is some symmetry that I'm missing or if I'm just entering it incorrectly in Mathematica, but I can't seem to get around the missing factor 2.
Relevant Equations
Polar coordinates in triple integral
The integral is$$\int_0^4dz\iint xyz~dxdy$$Constricted to the quarter circular disk $x^2+y^2=4$ in the first quadrant.
First I switched to polar coordinates and integrated the double integral by first writing it as:$$\int_0^4z~dz \int_0^\frac{\pi}2\int_0^2 \frac{1}{2}r^3sin(2\theta)~drd\theta$$
Which leads to:$$2\int_0^4z~dz \int_0^\frac{\pi}2 sin(2\theta) d\theta = 2\int_0^4z~dz = 16$$
I have tried to figure out what's missing and why it comes out as twice this in Mathematica and I feel the textbook is
scarce with descriptions. For example, we're asked to describe the region of integration and I think it's a quarter cylinder
with radius 2 and height 4 but can't find any support in the textbook for it.

I also can't understand why Mathematica gives the answer 16 when $\theta$ is integrated from $0$ to $\frac{\pi}4$
But that's not the quarter of the circle.
Btw, the code that I entered in Mathematica is: Integrate[x*y*z, {z, 0, 4}, {x, y} $\in$ Circle[{0, 0}, 2, {0, \[Pi]/2}]]

Any hints or pointers to what is missing in my reasoning?
Tnx.

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#### Mark44

Mentor
The integral is$$\int_0^4dz\iint xyz~dxdy$$Constricted to the quarter circular disk $x^2+y^2=4$ in the first quadrant.
First I switched to polar coordinates and integrated the double integral by first writing it as:$$\int_0^4z~dz \int_0^\frac{\pi}2\int_0^2 \frac{1}{2}r^3sin(2\theta)~drd\theta$$
Which leads to:$$2\int_0^4z~dz \int_0^\frac{\pi}2 sin(2\theta) d\theta = 2\int_0^4z~dz = 16$$
I have tried to figure out what's missing and why it comes out as twice this in Mathematica and I feel the textbook is
scarce with descriptions. For example, we're asked to describe the region of integration and I think it's a quarter cylinder
with radius 2 and height 4 but can't find any support in the textbook for it.

I also can't understand why Mathematica gives the answer 16 when $\theta$ is integrated from $0$ to $\frac{\pi}4$
But that's not the quarter of the circle.
Btw, the code that I entered in Mathematica is: Integrate[x*y*z, {z, 0, 4}, {x, y} $\in$ Circle[{0, 0}, 2, {0, \[Pi]/2}]]

Any hints or pointers to what is missing in my reasoning?
Tnx.
I get the same answer you did -- 16, and WolframAlpha agrees with this. See https://www.wolframalpha.com/input/?i=integrate+z+*+r^3+cos+t+sin+t+dr+dt+dz,+r=0..2,+t=0..pi/2,+z=0..4
My integral looks like this:
Code:
integrate z * r^3 cos t sin t dr dt dz, r=0..2, t=0..pi/2, z=0..4
My guess is that you misplaced a factor of 1/2 when you integrated $\sin(2\theta)$.

BTW, questions about integrals belong in the Calculus & Beyond section, not in the Precalc section you originally posted in.

#### Ray Vickson

Homework Helper
Dearly Missed
The integral is$$\int_0^4dz\iint xyz~dxdy$$Constricted to the quarter circular disk $x^2+y^2=4$ in the first quadrant.
First I switched to polar coordinates and integrated the double integral by first writing it as:$$\int_0^4z~dz \int_0^\frac{\pi}2\int_0^2 \frac{1}{2}r^3sin(2\theta)~drd\theta$$
Which leads to:$$2\int_0^4z~dz \int_0^\frac{\pi}2 sin(2\theta) d\theta = 2\int_0^4z~dz = 16$$
I have tried to figure out what's missing and why it comes out as twice this in Mathematica and I feel the textbook is
scarce with descriptions. For example, we're asked to describe the region of integration and I think it's a quarter cylinder
with radius 2 and height 4 but can't find any support in the textbook for it.

I also can't understand why Mathematica gives the answer 16 when $\theta$ is integrated from $0$ to $\frac{\pi}4$
But that's not the quarter of the circle.
Btw, the code that I entered in Mathematica is: Integrate[x*y*z, {z, 0, 4}, {x, y} $\in$ Circle[{0, 0}, 2, {0, \[Pi]/2}]]

Any hints or pointers to what is missing in my reasoning?
Tnx.
I get the same answer as you, using rectangular coordinates instead of polar.
$$I = \int_{z=0}^4 \, dz \int_{x=0}^2 \, dx \int_{y=0}^{\sqrt{4-x^2}} xyz \, dy \\ = \int_0^4 \, dz \int_0^2 \frac 1 2 z x (4-x^2) \, dx\\ = \int_0^4 2z \, dz = 16$$

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