Question

Help calculate the area of the graph of f(x,y) = 3/2

Answer 1

The total area = area ABED + area BCE
area ABED = \((\dfrac{1}{\sqrt2}-\dfrac{1}{2})(\dfrac{1}{2})=0.10355\)

Answer 2

You can determine the area with a double integral:
\[\int_{1/2}^1\int_{1/2}^{\sqrt x}dy~dx\]

Answer 3

yes, but I can't find out the right answer from it

Answer 4

After calculating, I got 0.116 for this part, adding with the area of rectangle, I got 0.219 while the answer from my prof is 0.2714
he said it is \(\dfrac{5}{8}-\dfrac{1}{2\sqrt 2}=0.2714\) and asked us find out the way to get that solution :(

Answer 5

oh, no, it is not your int

Answer 6

.2714 is wrong

Answer 7

B =(\sqrt 2/2, 1/2)

Answer 8

yes

Answer 9

\[\int\limits_{1}^{9}\]

Answer 10

hence, the int is
\[\int_{1/2}^1\int_{1/\sqrt 2}^{\sqrt y}3/2dx~dy\]

Answer 11

I check on wolfram and it has the same answer with me, how to get 0.2714??

Answer 12

http://www.wolframalpha.com/input/?i=%5Cint_%7B1%2F2%7D%5E%281%2Fsqrt%282%29%29%5Cint_%7B1%2F2%7D%5E%7B1%7D+%283%2F2%29dydx+%2B+%5Cint_%7B1%2Fsqrt%282%29%7D%5E1%5Cint_%7Bx%5E2%7D%5E%7B1%7D+%283%2F2%29dydx

Answer 13

you could also find the area and multiply by 3/2 as the function is a constant

Answer 14

Question: why do we have to take the first integral when it is a concrete rectangular?

Answer 15

You don't need to use double integrals if you don't want to as the function is constant

Answer 16

http://www.wolframalpha.com/input/?i=3%2F2*%5Cint_%281%2F2%29%5E1+%28sqrt%28y%29-1%2F2%29dy

Answer 17

\[\iint_R f(x, y) dA = \iint_R 3/2 dA = 3/2 (\text{Area})\]

Answer 18

You're integrating the function f(x,y) = 3/2 over the given region. This is a volume problem :