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11.2 mvh - Area between curves
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<h2>[11.2] - Area between curves
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<h4 style="background-color:pink"><a href="https://www.desmos.com/calculator/1bz7dhchuq">This desmos graph</a> has a better representation of the function $x=y^{\frac 14}$ than the plots below for b and c. Use the desmos one to estimate areas.</h4>

<p><img src="0g/12.0.areas.jpg">


<p class="ustme">a. 
$$\int_{x=0}^{3/4}\int_{y=x/4}^{x-x^2}dy\,dx$$
<span class="answer">=0.070</span>
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<p>b. Solving for $y$ by squaring both sides of $\sqrt{y}=x$ gives us $y=x^2$. 
$$\int_{x=0}^{1}\int_{y=x^2}^{\sqrt[4]{x}} dy\,dx$$
<span class="answer">=7/15</span><p>c. Now we want functions $x(y)$ of $y$. Solving $y=\sqrt x$ for $x$ gives
$x=y^2$ which is the function closest to the $y$ axis.
$$\int_{y=0}^{1/2}\int_{x=y^2}^{\sqrt[4]{y}} dx\,dy$$
<p>d. When solving for $x$, we take the square root of both sides of $y=(x-2)^2$.  The left side could be either $-\sqrt y$ or $+\sqrt y$.  The shape works out right if you choose the negative root, and then
$$
\int_{y=0}^1\int_{x=\sqrt y}^{2-\sqrt y} dy\,dx.$$
<span class="answer">=2/3</span><br><br><br>
2.) Think of $\sqrt{1-x^2-y^2}$ as the surface $z=\sqrt{1-x^2-y^2}$.  Then think of the argument of the integral as the difference between two functions, $z(x)-0$.... 
 
<p>And the integral can be written:
$$\int_{x=-1}^{1}\int_{y=-\sqrt{1-x^2}}^{+\sqrt{1-x^2}} \int_{z=0}^{\sqrt{1-x^2-y^2}}dz\,dy\,dx=\frac23 \pi$$
...because we are integrating $dV=dz\,dy\,dx$ over a region which is half of a sphere of radius 1!

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