areacalculusintegration
I am having some difficulty with this problem:
Find the area of the region enclosed by the curves.
$$4x+y^2=12,x=y.$$
So, the intersecting points I found were $-6$ and $2$.
$$\int\limits_{-6}^2 \frac{\sqrt{12 – y^2}}{4} dy – \int\limits_{-6}^2 y dy$$
I'm not sure how to solve this, any help please.
From Wolfram Alpha, we can sketch the curves to find the area of interest:
Note that we need to find the points of intersection: at $x = 0$ the lines $y = x, \;y = \frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \frac 1x$ intersect. At $x = 2,$ the lines $y = \frac 1x $ and $y = \frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$
Hint: Split it into two integrals. One where $2y+2x=6$ is under $y=3$ and one where $2y=4\sqrt x$ is under $y=3$. Then find where $2y=4\sqrt x$ and $2y+2x=6$ intersect, and that's the bound between the two integrals.
If you don't want two integrals, you can also integrate with respect to $y$, where the height is the difference between $2y=4\sqrt x$ and $2y+2x=6$ and the bounds are from $2$ to $3$.
Best Answer
In this case it is more convenient to consider
\begin{align} f_1(y)&=3-\tfrac14\,y^2 ,\\ f_2(y)&=y . \end{align}
\begin{align} S&= \int_{-6}^2\!\int_y^{3-\tfrac14\,y^2}\!dx\,dy = \left. 3y-\tfrac1{12}y^3-\tfrac12 y^2\right|_{-6}^2 =\tfrac{64}3 . \end{align}