# Volume bounded by two surfaces, what am I missing?

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1. Nov 5, 2015

### qq545282501

1. The problem statement, all variables and given/known data
Find the volume of the solid bounded by z=x^2+y^2 and z=8-x^2-y^2

2. Relevant equations
use double integral dydx

the text book divided the volume into 4 parts,

3. The attempt at a solution

f(x)= 8-x^2-y^2-(x^2+y^2)= 4-x^2-y^2

i use wolfram and got 8 pi, the correct answer is 16 pi from the text book.

I understood the text book version, I see why its dividing into 4 parts since its symmetric about x and y axis, but I dont understand why my method is half of the correct volume. obviously I need to multiply my double integral by 2, but I just don't get the picture, I am confused, am I just getting the volume of the lower paraboloid with my setup like that?

what if the upper paraboloid has different volume as the lower paraboloid, for example, the upper function could be z=4x^2+y^2, then multiplying by 2 just doesnt make sense since the top and bottom are not equal to each other.

can someone help me out, its been bothering me since yesterday.

Last edited by a moderator: Nov 5, 2015
2. Nov 5, 2015

### Staff: Mentor

Please don't use SIZE tags -- I have removed them from your post.

The above is incorrect. $8 - x^2 - y^2 - (x^2 + y^2) = 8 - 2x^2 - 2y^2$.

3. Nov 5, 2015

### qq545282501

omg, no wonder why! I was reading the text and just copied that down without even checking it. Thank you so much!!

4. Nov 5, 2015

### HallsofIvy

Staff Emeritus
Recheck the text. Possibly it was $8- x^2- y^2- (x^2+ y^2)= 2(4- x^2- y^2)$ and you missed the "2".

5. Nov 5, 2015

### Ray Vickson

It is a lot easier if you switch to polar coordinates for x and y; then you don't need to divide anything into parts, you just need to use the shell method.

6. Nov 5, 2015

### qq545282501

yes, you are right, i was not careful enough...

7. Nov 5, 2015

### qq545282501

yes, indeed. Though, i was trying to make sure I got the concept down in my mind.