- #1

Sparky_

- 227

- 4

## Homework Statement

Greetings:

My problem is solve using cylindrical coordinates:

[tex]

\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{0}^{y+2} xy dz dy dx

[/tex]

## Homework Equations

## The Attempt at a Solution

Using

[tex]

r^2=x^2 + y^2

x=r\cdot cos(\theta)

y=r\cdot sin(\theta)

[/tex]

I get:

[tex]

\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4} r^3 cos(\theta)sin(\theta)dz dr d\theta

[/tex]

[tex]

4\cdot \int_{0}^{2\pi} \int_{0}^{2} r^3 cos(\theta)sin(\theta) dr d\theta

[/tex]

[tex]

4\cdot \int_{0}^{2\pi} (r^4)/4 cos(\theta)sin(\theta) d\theta

[/tex]

Evaluated r from 0 to 2

[tex]

16\cdot \int_{0}^{2\pi} cos(\theta)sin(\theta) d\theta

[/tex]

[tex]

16/2 \cdot sin^2(\theta)

[/tex]

From 0 to [tex] 2\pi[/tex]

Gives 0.

OR! – and my question

[tex]

(16/2) \cdot 4 \cdot sin^2(\theta)

[/tex]

From 0 to [tex] \pi/2 [/tex]

Gives 32

I’m told the answer is 0. – Granted when you integrate a full rotation of a sinusoid you get 0 but the area under a sinusoid is not 0. -

But I thought to evaluate integrals of sin and cos because of the symmetry it was correct to evaluate the top half and then double it or evaluate the first quarter and then multiply by 4.

Can you clarify?

Thanks

Sparky_