Find integral area with respect to x

Question

The question is finding the area enclosed by the curves x=$y^2$ and x+2y=8 using both x and y integrals

Graph for reference Purple is x+2y=8, red is x=$y^2$

First I found the limits by letting x=8-2y. This gave the equation $y^2$+2y-8 which gave y=-4 and y=2.

Putting back into equation gives x=4 and x=16

I calculated with y integral being

$$\int_4^2 (8-2y)-y^2 \,$$ ( lower integral is -4, I can't seem to express)

Now I would like to ask how would you calculate using the x integral

I had a think about this and got

but the answers x and y integrals are different so I think theres a mistake somewhere but I don't know what I did wrong.

Answer 1

Your first integral is slightly incorrect. The correct setup and solution would be: $$A=\int_{-4}^{2} \left(8-2y-y^2\right) \,dy=36\ \text{sq. units}.$$

If you now want to do integration with respect to $x$, begin by expressing your two functions as functions of $x$:

$$ y=\pm\sqrt{x},\\ y=-\frac{x}{2}+4. $$

And then find where the curves intersect each other and the $x$-axis. That happens at $x=4$, $x=8$ and $x=16$.

Now you can do the integration, keeping in mind that area below the $x$-axis counts as negative:

$$ A=\int_{0}^{4}\sqrt{x}\,dx+\int_{4}^{8}\left(-\frac{x}{2}+4\right)\,dx -\int_{0}^{16}\left(-\sqrt{x}\right)\,dx +\int_{8}^{16}\left(-\frac{x}{2}+4\right)\,dx=\\ \int_{0}^{4}\sqrt{x}\,dx + \int_{0}^{16}\sqrt{x}\,dx +\int_{4}^{16}\left(-\frac{x}{2}+4\right)\,dx= 36\ \text{sq. units}. $$

Answer 2

The first should be $$\int\limits_{-4}^2(8-2y-y^2)dy.$$

The second should be $$\int\limits_0^4\left(\sqrt{4-\frac{x}{2}}-\left(-\sqrt{4-\frac{x}{2}}\right)\right)dx+\int\limits_4^{16}\left(4-\frac{x}{2}-\left(\sqrt{4-\frac{x}{2}}\right)\right)dx$$

Answer 3

Reverse the $x$ and the $y$ coordinate axis (which amounts to a symmetry, an operation that doesn't change the absolute value of an area ; see figure below), giving equations

$$y=x^2 \ \text{and} \ y+2x=8 \ \ \ \iff \ \ \ y=x^2 \ \text{and} \ y=-2x+8 \tag{1}$$

In this way, taking into account the fact that the intersection points of the new curves are $(x,y)=(\color{red}{-4},16)$ and $(x,y)=(\color{red}{2},4)$, it remains to compute (using the 2nd form of equations (1)) the simple integral:

$$\int_{\color{red}{-4}}^{\color{red}{2}}((-2x+8)-x^2)dx=\left[-x^2+8x-\tfrac{x^3}{3}\right]_{-4}^2=36.$$