# SOLVEDJust need someone to verify integral, please

#### yeKciM

I got this one, and tree times I check it, and found few mistakes, and I corrected them all, but now I'm not so sure if i got it all corrected. So if anybody could check it and tell me where do I do it wrong, I'll bee grateful integral is :

$$\displaystyle \int\int \ln{x^2+y^2}\,dx\,dy$$

by the region D defined by:

$$\displaystyle x^2+y^2=e^2$$
$$\displaystyle x^2+y^2=e^4$$

so I used :

$$\displaystyle x=\rho\cos\varphi$$
$$\displaystyle y=\rho\sin\varphi$$
$$\displaystyle J=\rho$$

and I got that my limits now are for $$\displaystyle \rho$$ from $$\displaystyle e$$ to $$\displaystyle e^2$$, and for $$\displaystyle \varphi$$ from $$\displaystyle 0$$ to $$\displaystyle 2\pi$$.

now my integral is:

$$\displaystyle I=\displaystyle\int_0^{2\pi} \,d\varphi \int_e ^{e^2} \ln(\rho^2) \rho \,d\rho = 2\int_0^{2\pi} \,d\varphi \int_e ^{e^2} \ln(\rho) \rho \,d\rho$$

so I now do one first...

$$\displaystyle I_1= \displaystyle \int_e ^{e^2} \ln(\rho)\rho\,d\rho$$

$$\displaystyle u=\ln(\rho), \,du=\frac{1}{\rho} \,d\rho$$

$$\displaystyle \,dv=\rho\,d\rho , v=\frac{\rho^2}{2}$$

so I get that one like this...

$$\displaystyle \displaystyle I_1=\frac{\rho^2}{2}\ln(\rho) \mid _e ^{e^2} - \int_e ^{e^2} \frac {\rho^2}{2} \frac{1}{\rho} \,d\rho$$
$$\displaystyle \displaystyle I_1=e^4-\frac{e^2}{2}-\frac{1}{4} (e^4-e^2)=\frac{3e^4}{4}-\frac{e^2}{4}$$

so my complete integral is :

$$\displaystyle \displaystyle I=2\int_0 ^{2\pi} [\frac{3e^4}{4}-\frac{e^2}{4}] \,d\varphi= \frac {3e^4-e^2}{2} 2\pi = \pi (3e^4-e^2)$$

(Thinking)(Worried)(Thinking)

#### Ackbeet

MHF Hall of Honor
Looks good to me.

#### yeKciM

Thanks Ackbeet I got one more question... if i have integral, and have to compute volume of figure defined with $$\displaystyle x^2+y^2+z^2=8$$ and $$\displaystyle x^2+y^2=z^2 , z\ge0$$, but it's explicitly said to use double integral...
is it correct to do this:

$$\displaystyle \displaystyle V=\int\int_D \sqrt{8-x^2-y^2} \,dx\,dy-\int\int_D (x^2+y^2) \,dx\,dy$$

after change of coordinate system ... i get result ... my question is : is this only way to solve this ... to do double integral of cylinder and then subtract double integral from cone from it? or there is some another way to do this type of integration #### Ackbeet

MHF Hall of Honor
I'm not sure I've wrapped my head around the exact figure of which you're trying to find the volume. You've got a sphere centered at the origin of radius $$\displaystyle 2\sqrt{2},$$ and you've got a cone opening upwards only. Are you trying to find the volume of the intersection of those two shapes, or are you trying to find the volume of the sphere less the volume of the cone inside the sphere?

Either way, I'm puzzled how the restriction on doing this with double integrals only will enhance instruction! You really don't need any integration to find the volume of the sphere - its volume is a well-known formula (which can be derived, it is true, by using triple integration).

One tricky aspect of this problem: if your problem is to find the volume of the sphere minus the volume of the cone inside it, I would clarify whether the top of the cone (the circular region) is flat, or follows the contour of the sphere. That will make a difference in your final result!

#### yeKciM

hmmmm... task is to show how to do double integrals in $$\displaystyle R^3$$ space (compound volume) and yes it's figure that creates cone inside sphere... like cornet with one ice cream ball in it (chocolate maybe)    it's more easy for me to to any volume with triple integration, but I look at past exams from another generations and on every single exam it was at least one of problem with this type of integration  P.S. radius of sphere is $$\displaystyle \sqrt8$$ and their intersection at radius 2 but that doesn't matter right now just, am I correct with (if that's figure that I'm compounding volume) when i do [volume of cylinder - volume of cone ] so i can get volume of that cornet with one ice cream ball inside (on top that is ) #### Ackbeet

MHF Hall of Honor
Ok. I understand the shape the volume of which you're trying to find. And I think I follow your comments - they're a little piece-meal and disconnected.

I think I'd use cylindrical coordinates for this problem. That takes advantage of the azimuthal symmetry of the problem. I think I'd find the volume of the intersection of the sphere and the cylinder of radius 2 (that's the "radius 2" you mentioned) that's above the z = 0 plane, and then subtract from that the volume under the cone that's still inside that same cylinder. That is, I'd do the following:

$$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\sqrt{8-r^{2}}\,r\,dr\,d\varphi-\int_{0}^{2\pi}\int_{0}^{2}r^{2}\,dr\,d\varphi.$$

I think I've set that up properly. Would you agree? Does it make sense?

• yeKciM

#### yeKciM

yes thanks i did that, but I wasn't so sure that is right thing to do bu i didn't see any another way of doing the task with double integral thank you very much again #### Ackbeet

MHF Hall of Honor
You're very welcome.

Here's some checks you can do to see if your final answer makes sense:

The volume computed should be slightly greater than the cone described by $$\displaystyle r^{2}=z^{2}, 0\le r\le 2,$$ which you can compute easily. The volume should be less than the first integral I mentioned, representing the intersection of the sphere with the cylinder of radius 2 above the z = 0 plane.

Finally, I doubt seriously whether anyone would prevent you from computing the volume using a triple integral to check your work. There, I would set it up as follows (using $$\displaystyle \theta$$ as the polar angle and $$\displaystyle \varphi$$ as the azimuthal angle):

$$\displaystyle V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{2\sqrt{2}}\rho^{2}\sin(\theta)\,d\rho\,d\theta\,d\varphi.$$

P.S. In post # 5, you mentioned that the radius is $$\displaystyle \sqrt{8}.$$ That's true. It's also true that $$\displaystyle \sqrt{8}=2\sqrt{2}.$$ (Wink)

#### yeKciM

yes i know just can't believe that i wrote down more math, solving it with triple integral hehehehe setting up limits took a while.. a lot of things I knew it right away but i started to write everything, just in case thank u again for everything

P.S. In post # 5, you mentioned that the radius is $$\displaystyle \sqrt{8}.$$ That's true. It's also true that $$\displaystyle \sqrt{8}=2\sqrt{2}.$$ (Wink)
sorry, i didn't mean to say that you are wrong saying it's $$\displaystyle 2\sqrt2$$... I was justs saying what i got, with accent on that intersection didn't be so sure it's radius 2 (Bow) (Bow) (Bow)

#### Ackbeet

MHF Hall of Honor
You're welcome, and no offense taken.

Have a good one!