# Soliving a double integral, need help

1. Oct 3, 2004

### babbagee

Ok the question is

find the volume of the region inside the surface z = x2 + y2 and between z = 0 and z = 10.

Ok i have already found the limits of integration but i am having a hard time calculating the integral.

The limits are $$-{\sqrt{10-x^2}$$ <= y <= $${\sqrt{10-x^2}$$
$$-{\sqrt{10}$$ <= x <= $${\sqrt{10}$$

I am integrating with respect to y first and then x.

The problem is after i integrate with respect to y i get a really hard integral which i tried soliving by trignometric substitution but i just made more complicated.

Can someone help me out please

thanks

2. Oct 3, 2004

### arildno

Here's a trick if you didn't manage the trig-substitution:
Volume inside parabolic=Volume of cylinder minus volume under parabolic.
The easiest way to do the last volume, is to change into polar coordinates.

3. Oct 3, 2004

### babbagee

I think they are asking for the volume under the paraboloid because i integrated once and then i entered the rest of it into the calculator and got the same answer as the back of the book. So i dont think they are asking for the volume above the paraboloid even though it sounds like it. That is what i thought at first. I dont think i need to convert into polor coordinates becuase the book has not gone into that much detail about converting into different coordinate systems, that is the next chapter. My limits are based on the volume under the paraboloid. I was at first thing of taking the volume under the paraboloid and subtracting it from the the cube that the paraboloid is enclosed in, its the same concept as the cylinder. But i dont think this problem wants me to do all that. Is there any way i can calculate the second integral. and by the way the answer in the back of the book as 50pi, and my calculater gave an answer of 157.0796 which is 50pi.

Thanks

4. Oct 3, 2004

### arildno

I don't think the trig-sub is really that difficult..
(Haven't checked, though)

5. Dec 11, 2008

### lopared

ok, this is old but still a good example of cylindrical coordinates.
The rectangular integral to solve this would be

$$\int_{-\sqrt{10}}^{\sqrt{10}} \int_{-\sqrt{10-x^2}}^{\sqrt{10-x^2}} (x^2+y^2) dydx}$$

which if you plug into a computer solver is no problem, but unfortunately you will need another method to solve this integral by hand.
First off, why do you convert to polar coordinates here? Well if you think about the projection(or a shadow) of the paraboloid
on the xy-plane then you can see a circle with radius of $${\sqrt{10}}$$ and what better to represent circles than polar coordinate.
If we where dealing with a function in $$R^2$$ we know that $$x=rcos{\theta}; y=rsin{\theta}$$, which is true in $$R^3$$ and we just leave $$z=z$$.
$$dA$$ will become $$rdrd{\theta}$$ (this can be explained if you think of how polar coordinates work, infinitesimally small blocks of a circle can be
thought of as a normal rectangle with sides $$rdr$$ and $$d{\theta}$$ making the area $$rdrd{\theta})$$ so $$dA=dxdy=rdrd{\theta}$$ also notice that
$$x^2+y^2=r^2(cos^2{\theta}+sin^2{\theta})=r^2$$. We know that $$r$$ bounds are $$0\underline{<}r\underline{<}{\sqrt{10}}$$ and $$0\underline{<}{\theta}\underline{<}2{\pi}$$ so the new integral to solve is
$$\int_{0}^{2{\pi}} \int_{0}^{\sqrt{10}} r^2 rdrd{\theta} = \int_{0}^{2{\pi}} \int_{0}^{\sqrt{10}} r^3 drd{\theta} = \int_{0}^{2{\pi}} d{\theta} \int_{0}^{\sqrt{10}} r^3 dr = \frac{2{\pi}}{4}\sqrt{10^4} = \frac{200{\pi}}{4} = 50{\pi}$$

I hope this helps anyone that may be having similar problems with this type of problem.