Question

question to come ..

Answer 1

find the volume between the surfaces \[z = 2x ^{2} + y ^{2} + 12\] and \[z = x ^{2} + y ^{2} + 8, and \] over the triangle with vertices (0,0), (1,0), (1,2)

Answer 2

my solutions guide gave me a different answer than what i got .. if anyone does it lemme know :)

Answer 3

So if we call the equations of the two surfaces f2(x,y) and f1 respectively, we want
\[\int\limits_T (f_2 - f_1) dx dy\]
where t is the triangle.
The only problem is the integration limits. I'd do y = 0 to 2x and then x = 0 to 1.
I get 17/2. What's the book say?

Answer 4

i did it your way already and got the right answer. it is 9/2. but when i do it from y = 0 to 2 and x = 0 to .5y, i get something else.

Answer 5

Actually if you use those set of limits you just described, that is the triangle with vertices (0,0), (0,2) and (1,2)

Answer 6

how could that be true? an integral does area under the curve.

Answer 7

If you integrate x first with those limits, you are integrating from the line x = 0 to the line x = y/2.

Answer 8

By the way, where I am screwing up in the integral I proposed? Isn't f2 - f1 = x^2 + 4

Answer 9

So then\[\int\limits_0^1 \int\limits_0^{2x} (x^2 + 4) dy dx\]

Answer 10

\[= \int\limits_0^1 (2x^3 + 16x) dx = 17/2.\]

Answer 11

Oh, I see it: 2x^3 + 8x. Got it. Hence 9/2

Answer 12

Right. Anyway ... I stand by my argument for your other limits: you're integrating over a different triangle.

Answer 13

so how would you set it up with my triangle

Answer 14

It would be x from y/2 to 1. And then y from 0 to 1.

Answer 15

Draw the diagram ... I wish were were at a blackboard; it's too hard to draw here.

Answer 16

y goes from 0 to 2

Answer 17

Ah yes, sorry.