I'm trying to calculate the following integral:
$$I=\iint_{D}(x-\frac{7}{3})y \; dxdy,$$ where $D$ is the area limited by the curves $\frac{7}{3}-2y-x=0$ and $x=\frac{1}{3}y^{2}$.
I'll insert a picture of my solution. I would love to hear if you can find somewhere where I've done something crucially wrong. I just can't get the right answer at the moment. I'm not expecting you to control all the constants and do the whole calculation since it's very time consuming.
Thanks!
Your integral set up is correct but there is a mistake in your work.
$\displaystyle \small I = \int_{-7}^1 y \left[\int_{y^2/3}^{(7/3 - 2y)} \left (x - \frac 73 \right) dx \right] dy = \int_{-7}^1 y \left[\frac {x^2}2 - \frac {7x}3\right]_{y^2/3}^{7/3-2y} dy$
$\small \displaystyle = \int_{-7}^1 y \left(\frac 12 \left(\color{blue}{\frac {49}9} - \frac {28y}{3} + 4y^2\right) - \frac73 \left(\frac73-2y\right) - \frac{y^4}{18} + \frac{7y^2}{9} \right) dy$
See the part highlighted in blue. You instead have $ \displaystyle \frac{21}{9}$. Rest of the work is correct and you get an answer of $-512$ after fixing the mistake.
Change of variables can simplify the work.
$u = x - \frac 73, v = y^2$ does simplify the integrand as the Jacobian of transformation is $\frac{1}{|y|}$ but you need to be careful about the sign.
Even $X = x - \frac{7}{3}$ simplifies the work to an extent. The equations of the curves are then,
$2y + X = 0, y^2 = 3X + 7$
The integral becomes,
$ \small \displaystyle I = \int_{-7}^1 y \left[\int_{(y^2-7)/3}^{-2y} X ~ dX \right] dy$
$ \displaystyle \small = \frac 1 {18} \int_{-7}^1 (50 y^3 - y^5 - 49y) ~dy$