- #1
TheMercury79
- 24
- 5
- Homework 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.