Integral with sinsoid 0 or non-zero help?

Homework Statement

Greetings:

My problem is solve using cylindrical coordinates:

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

The Attempt at a Solution

Using

$$r^2=x^2 + y^2 x=r\cdot cos(\theta) y=r\cdot sin(\theta)$$

I get:

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

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

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

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

$$16/2 \cdot sin^2(\theta)$$
From 0 to $$2\pi$$

Gives 0.

OR!! – and my question

$$(16/2) \cdot 4 \cdot sin^2(\theta)$$
From 0 to $$\pi/2$$

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_

The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Just do the integral 0 to 2pi. There is no rule that says integrate the first quadrant and multiply by 4. BTW your dz integration is wrong as well.

Thanks

Can I ask 2 follow-up questions

1)
Regarding my dz error -

Is the error in setting the integral up or in the actual integration?

integrate dz from 0 to 4 (no other terms within the integral are a function of z) gives "z" from 0 to 4 - which gives "4"

Is this not correct?

2)
Regarding the "no-rule that says take the first quadrant and multiply by 4" - I agree I've never heard of a "rule" for this - I've run into problems that (I think) required taking advantage of symmetry.

Can we discuss that a little more - to help me see it?

I think I recall when integrating a sinusoidal function over a full cycle the integral could yield something like 1 - 1 - but when taking the first half of the curve and then multiply it by 2 you get a non-zero "area under the curve."

I recall working a different problem in which I had to go from 0 to pi/2 and then muliply by 4 - going from 0 to 2pi or 0 to pi did not work.

Thoughts?

Thanks so much
-Sparky

Dick
Homework Helper
1) You are supposed to integrate xy from 0 to y+2 dz. You can't just replace the upper limit by 4. 2) Sure, symmetry is handy in some problems. But if you integrate sin(theta)*cos(theta) from 0 to pi/2 you get you get 1/2. If you integrate it from 0 to 2pi, as you are supposed to do, you get 0. This isn't an 'area' problem where you are supposed to count regions below the x-axis as positive.

Thanks

I think I'm good with this now - I'll correct my dz integral setup and think about my limits (0 to 2pi)

I may have another question but let me put some thought into this now.

Thanks Again!!

can you check my work - have I corrected my limits on z correctly?
Is the integration correct?

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

Using:

$$r^2=x^2 + y^2 x=r\cdot cos(\theta) y=r\cdot sin(\theta)$$

The Z goes from: 0 to $$rsin\theta + 2$$

So,

$$\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{rsin(\theta) +2} r*r cos(\theta)*rsin(\theta)dz dr d\theta$$

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

$$\int_{0}^{2\pi} \int_{0}^{2} Z r^3 cos(\theta)sin(\theta) dr d\theta$$
Evaluate – Z from 0 to $$rsin(\theta) + 2$$

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

$$\int_{0}^{2\pi} \frac {r^5}{5} sin^2(\theta) cos(\theta) + \frac{2r^4}{4}sin(\theta)cos(\theta) dr d\theta$$
Evaluate r from 0 to 2

$$= \frac{32}{5}\frac{sin^3(\theta)}{3} + (8) \frac{sin^2(\theta)}{2}$$
Evaluate $$\theta$$ from 0 to $$2\pi$$

= 0

Thanks!!

Dick
Homework Helper
Looks ok to me.

Dick (and others) –

I have a question regarding handling symmetry and changing the limits of integration – specifically the limits on $$\theta$$ in polar coordinates. Meaning, of there is symmetry I’ve seen the limits of a polar integration changed FROM 0 to $$2 \pi$$TO 0 to $$\pi$$and the integral is multiplied by 2 or an integral is multiplied by 4 and the limit of integration goes to $$\pi/2$$.

Dick – above you tell me to integrate from 0 to $$2\pi$$.

Below are 2 examples: 1) the $$\theta$$ limits are from 0 to $$2\pi$$ and 2) the integral is multiplied by 2 and the limits are from 0 to $$\pi/2$$.

The main question why does the 1st problem not use symmetry and keeps the limits of integration from 0 to $$2\pi$$while the 2nd problem uses “2 times the integral taken from 0 to $$\pi/2$$.”?

What process do I use / what line of analysis do I apply to determine how to set my limits?

Problem 1: (the one you helped me with Dick)

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

Using:

$$r^2=x^2 + y^2 x=r\cdot cos(\theta) y=r\cdot sin(\theta)$$

The Z goes from: 0 to $$rsin\theta + 2$$

So,

$$\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{rsin(\theta) +2} r*r cos(\theta)*rsin(\theta)dz dr d\theta$$

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

$$\int_{0}^{2\pi} \int_{0}^{2} Z r^3 cos(\theta)sin(\theta) dr d\theta$$
Evaluate – Z from 0 to $$rsin(\theta) + 2$$

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

$$\int_{0}^{2\pi} \frac {r^5}{5} sin^2(\theta) cos(\theta) + \frac{2r^4}{4}sin(\theta)cos(\theta) dr d\theta$$
Evaluate r from 0 to 2

$$= \frac{32}{5}\frac{sin^3(\theta)}{3} = (8) \frac{sin^2(\theta)}{2}$$
Evaluate $$\theta$$ from 0 to $$2\pi$$

= 0

Part of the question is notice the limits on theta go from 0 to 2$$pi$$
2)
Next problem:

This example is “part” of a problem stating: “Find the volume of the solid remaining when the region inside the cylinder $$r = 3sin(\theta)$$ is removed from the solid bounded by $$x^2 + y+2 + z^2 = 9$$.”

This is why I have the limits on r and z I do in the example below.

$$\int \int \int r dz dr d\theta r = 3sin(\theta)$$

$$\int_{0}^{2\pi} \int_{0}^{3sin(\theta)} \int_{- \sqrt{9-r^2}}^{\sqrt{9-r^2}} r dz dr d\theta r = 3sin(\theta)$$

$$2\int_{0}^{2\pi} \int_{0}^{3sin(\theta)} \int_{0}^{\sqrt{9-r^2}} r dz dr d\theta r = 3sin(\theta)$$

below: changing the limits on $$\theta$$ to 0 to $$\pi/2$$ and multiplying by 2
$$(2)(2)\int_{0}^{\pi/2} \int_{0}^{3sin(\theta)} \int_{0}^{\sqrt{9-r^2}} r dz dr d\theta r = 3sin(\theta)$$

$$(4)\int_{0}^{\pi/2} \int_{0}^{3sin(\theta)} rz dr d\theta r = 3sin(\theta)$$
Limits on z: 0 to $$\sqrt{9-r^2}$$

$$(4)\int_{0}^{\pi/2} \int_{0}^{3sin(\theta)} r\sqrt{9-r^2} dr d\theta$$

$$(-2)(2/3)\int_{0}^{\pi/2} (9-r^2)^(3/2) d\theta$$

$$(-4/3)\int_{0}^{\pi/2} (9-9sin^2(\theta))^(3/2)-(9)^(3/2) d\theta$$

$$=(-4/3)\int_{0}^{\pi/2} 27(1-sin^2(\theta))^(3/2)-(27) d\theta$$

$$=(-4(27)/3)\int_{0}^{\pi/2} cos^3(\theta) -1 d\theta$$

$$=(-4(27)/3)\int_{0}^{\pi/2} cos(\theta)(1-sin^2(\theta) -1 d\theta$$

$$=(-4(27)/3)\int 1-u^2 d\theta + 4(27)/3)\int d\theta$$

$$=(-4(27)/3)[u-(u^3)/3 + 4(9)(\pi)/(2)$$

$$=(-4(27)/3)[sin(\theta)-(sin^3(\theta))/3 + 18\pi$$
Limits: 0 to $$\pi/2$$

$$= -4(9)[1-(1/3) + 18\pi$$

$$= -24+ 18\pi$$

In problem 1 – limits on $$\theta$$ is from 0 to $$2\pi$$.

In problem 2 – limits on $$\theta$$ is from 0 to $$(\pi)/2$$ and multiply the integral by 2.

How do I examine a problem to decide how to clear this up?

Thanks
-Sparky_

Last edited: