Double Integral question

[kankerfist]

If R is the region in the xy-plane that is below the parabola \(\displaystyle y = x^2\), above the x-axis, and for which \(\displaystyle 0 \le x \le 2\), compute the integral:

\(\displaystyle \int {} \int {(x + y)dA}\)

a)by integrating y first, then x
b)by integrating x first then y

I know that if I set this up correctly, my answer to a) should equal b), but they are not. I am not sure which one I am setting up wrong, or if I am setting them both up wrong. This is my integral of y first then x:

\(\displaystyle \int\limits_0^2 {\int\limits_0^{x^2 } {(x + y)dydx} }\)

which yields an answer of 36/5, and this is my integral of x first, then y:

\(\displaystyle \int\limits_0^4 {\int\limits_{ - \sqrt y }^{\sqrt y } {(x + y)dxdy} }\)

which yields an answer of 128/5. Can someone point out which double integral I incorrectly set up, or if I set them both up wrong? Thanks a lot!

 

[pka]

Try \(\displaystyle \L
\int_0^4 {\int_{\sqrt y }^2 {(x + y)dxdy} }\)

 

[kankerfist]

Ah that works! How did you decide that x would integrate from y^(1/2) to 2?

 

[pka]

Ah that works! How did you decide that x would integrate from y^(1/2) to 2?

Draw the graph Shade the region.
Think about how x would vary.

 

[soroban]

Hello, kankerfist!

If R is the region in the xy-plane that is below the parabola \(\displaystyle y = x^2\), above the x-axis,
and for which \(\displaystyle 0 \le x \le 2\), compute the integral:

\(\displaystyle \L\int\int(x + y)\,dA\)

a) by integrating y first, then x
b) by integrating x first, then y

I'll do what pka suggested . . .

      |
      |       * (2,4) 
      |       :
      |      *:
      |     *::
      |   *::::
    --*-------+--
      |       2

\(\displaystyle \text{a) }y\text{ goes from }y\,=\,0\text{ to }y\,=\,x^2\)

\(\displaystyle \;\;\;\text{Then }x\text{ goes from }x\,=\,0\text{ to }x\,=\,2\)

\(\displaystyle \;\;\;\text{We have: }\L\,\int^{\;\;\;2}_0\int^{\;\;\;x^2}_0(x\,+\,y)\,dy\,dx\)


\(\displaystyle \text{b) }x\text{ goes from }x\,=\,\sqrt{y}\text{ to }x\,=\,2\)

\(\displaystyle \;\;\;\text{Then }y\text{ goes from }y\,=\,0\text{ to }y\,=\,4\)

\(\displaystyle \;\;\;\text{We have: }\L\,\int^{\;\;\;4}_0\int^{\;\;\;2}_{\sqrt{y}}(x\,+\,y)\,dx\,dy\)