Volume bounded by two surfaces, what am I missing?

[qq545282501]

Homework Statement

Find the volume of the solid bounded by z=x^2+y^2 and z=8-x^2-y^2

Homework Equations

use double integral dydx

the textbook divided the volume into 4 parts,

The Attempt at a Solution

[/B]
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 textbook.

I understood the textbook version, I see why its dividing into 4 parts since its symmetric about x and y axis, but I don't 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 doesn't make sense since the top and bottom are not equal to each other. can someone help me out, its been bothering me since yesterday.

 

[Mark44]

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

Homework Statement

Find the volume of the solid bounded by z=x^2+y^2 and z=8-x^2-y^2

Homework Equations

use double integral dydx

the textbook divided the volume into 4 parts,

View attachment 91365

The Attempt at a Solution

[/B]
f(x)= 8-x^2-y^2-(x^2+y^2)= 4-x^2-y^2

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

View attachment 91364
i use wolfram and got 8 pi, the correct answer is 16 pi from the textbook.

I understood the textbook version, I see why its dividing into 4 parts since its symmetric about x and y axis, but I don't 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 doesn't make sense since the top and bottom are not equal to each other. can someone help me out, its been bothering me since yesterday.

 

[qq545282501]

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

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

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

 

[HallsofIvy]

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".

 

[Ray Vickson]

Homework Statement

Find the volume of the solid bounded by z=x^2+y^2 and z=8-x^2-y^2

Homework Equations

use double integral dydx

the textbook divided the volume into 4 parts,

View attachment 91365

The Attempt at a Solution

[/B]
f(x)= 8-x^2-y^2-(x^2+y^2)= 4-x^2-y^2

View attachment 91364
i use wolfram and got 8 pi, the correct answer is 16 pi from the textbook.

I understood the textbook version, I see why its dividing into 4 parts since its symmetric about x and y axis, but I don't 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 doesn't make sense since the top and bottom are not equal to each other. can someone help me out, its been bothering me since yesterday.

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.

 

[qq545282501]

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".

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

 

[qq545282501]

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.

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