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<title>
11.2 mvh - Double integrals practice
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<h2>
[11.2] - Double Integrals - Practice
</h2>
<ol>
<li>$\int_0^2\int_0^2(x^2-y^2)\,dy\,dx$
<br><br><br><br><br><br>

<li>$\int_0^{\frac{\pi}{4}}\int_{y=0}^{\frac{\pi}{2}}\cos(2x+y)\,dy\,dx$
<br>Hint: According to one of the <a href="http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html">Trig addition formulas</a>, $\cos(\alpha+\beta)=
\cos(\alpha)\cos(\beta)  - \sin(\alpha)\sin(\beta)$
<br><br><br><br><br><br>

<li>$\int_0^2\int_{y=1}^3x^3y\,dy\,dx$
<br><br><br><br><br><br>

<li>$\int_{-1}^{1}\int_{y=0}^{\pi}x^2\sin y\,dy\,dx$
<br>
<div class="justme">
$$\begineq
\iint &=\int_{-1}^{1}{\color{blue}\int_{y=0}^{\pi}x^2\sin y\,dy}\,dx\\
&=\int_{-1}^{1}{\color{blue}x^2\int_{y=0}^{\pi}\sin y\,dy}\,dx

=\int_{-1}^{1}{\color{blue}x^2\Big(-\cos y\Big)_0^{y=\pi}} \,dx\\

&=\int_{-1}^{1}{\color{blue}x^2\left(-(-1) - (-(1))\right)}\,dx \\

&={\color{blue}2}\left(\frac13x^3\right)_{-1}^{x=+1}\,dx = {\color{red}\frac43}.\\

\endeq
$$
</div><br><br><br><br><br>

<li>$\int_0^2\int_{y=x^2}^{2x}(x^2+2y)\,dy\,dx$
<br><div class="justme">
$$\begineq
\iint &= \int_0^2{\color{blue}\int_{y=x^2}^{2x}(x^2+2y)\,dy}\,dx \\
 &= \int_0^2{\color{blue}\Big(x^2y+y^2\Big)^{y=2x}_{x^2}}\,dx\\
 &= \int_0^2{\color{blue}\Big(x^2\cdot 2x+(2x)^2-x^2\cdot x^2-(x^2)^2\Big)}
 \,dx \\
 &= \int_0^2{\color{blue}\Big(2x^3+4x^2-2x^4\Big)} \,dx \\
  &= \Big(\frac24x^4+\frac43x^3-\frac25x^5\Big)_0^{x=2} \\
    &= \frac12(16)\frac{15}{15}+\frac43(8)\frac55-\frac25(32)\frac33 =\frac{120}{15}+\frac{160}{15}-\frac{192}{15}\\
    \\&=\color{red}\frac{88}{15}\approx 5.867
\endeq
$$
</div>
<br><br><br><br><br>

<li>$\int_0^3\int_{y=0}^{9-x^2}4x\,dy\,dx$
<br><br>


</ol>
<h3 style="page-break-before:always"></h3>


<a href="#1" onclick="toggle_visibility('a1');">show / hide</a> answers:
<div id="a1" class="answer toggle">

Answers: 1.) 0 &nbsp; &nbsp; 2.) 0 &nbsp; &nbsp; 
3.) 16 &nbsp; &nbsp; 
4.) $\frac 43$ &nbsp; &nbsp; 
5.) $\frac{88}{15}$ &nbsp; &nbsp; 
6.) 81
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