- 8

- 4

- 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.

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.