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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.

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3 Answers

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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}

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Let’s add up tiny rectangles of widths along the $x$-direction and infinitesimal height $dy$.

The two curves are defined as $x=y$ and $x=\frac14(12-y^2)$. Therefore, the width of each rectangle is $$\frac{12-y^2}{4}-y$$

That gives the infinitesimal area $$dA=\left(\frac{12-y^2}{4}-y\right)\,dy$$ We can see that we’re adding up rectangles from $y=-6$ to $y=2$, so we just have to evaluate

$$\begin{align} A &=\int_{-6}^{2} \left(\frac{12-y^2}{4}-y\right) \, dy \\ &= \frac14 \int_{-6}^{2}\left(12-y^2\right)\, dy -\int_{-6}^{2} y\,dy \\ &= \frac14 \left(12y-\frac13y^3\middle)\right|_{-6}^{2} - \left( \frac12y^2\middle)\right|_{-6}^{2} \end{align}$$

Can you handle it from there?

You were on track except for taking the square root.

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The area below the $x-$ axis is \begin{eqnarray*} \int_{-6}^{3} \sqrt{12-4x} dx - \frac{1}{2}6 \times 6. \end{eqnarray*} The area above the $x-$ axis is \begin{eqnarray*} \int_{2}^{3} \sqrt{12-4x} dx + \frac{1}{2}2 \times 2. \end{eqnarray*}

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