<span>up vote2down voteacceptedI think your limits of integration are incorrect. If you substitute <span><span>y=4</span><span>y=4</span></span> into <span><span><span>y2</span>−<span>x2</span>=9</span><span><span>y2</span>−<span>x2</span>=9</span></span>, you find that <span><span>x=±<span>7–√</span></span><span>x=±7</span></span>. Therefore, the two curves intersect at <span><span>x=±<span>7–√</span></span><span>x=±7</span></span>. By washer method, we have:<span><span>V<span><span>=π<span>∫<span>7√</span><span>−<span>7√</span></span></span>(4<span>)2</span>−(<span><span><span>x2</span>+9</span><span>−−−−−</span>√</span><span>)2</span>dx</span><span>=2π<span>∫<span>7√</span>0</span>16−(<span>x2</span>+9)dx</span><span>=2π<span>∫<span>7√</span>0</span>7−<span>x2</span>dx</span><span>=2π<span><span>[<span>7x−<span>13</span><span>x3</span></span>]</span><span>7√</span>0</span></span><span>=2π<span>(<span><span>14<span>7–√</span></span>3</span>)</span></span><span>=<span><span>28π<span>7–√</span></span>3</span></span></span></span><span><span>V<span>=π<span>∫<span>−7</span>7</span>(4<span>)2</span>−(<span><span>x2</span>+9</span><span>)2</span>dx</span></span><span>=2π<span>∫07</span>16−(<span>x2</span>+9)dx</span><span>=2π<span>∫07</span>7−<span>x2</span>dx</span><span>=2π<span><span>[<span>7x−<span>13</span><span>x3</span></span>]</span>07</span></span><span>=2π<span>(<span><span>147</span>3</span>)</span></span><span>=<span><span>28π7</span>3</span></span></span></span>And just for fun, let's try the shell method. Here, we have no choice but to find the volume obtained by revolving just the part of the region in the first quadrant, and doubling it.<span><span>VVVVVV</span><span><span>=2×2π<span>∫43</span>y<span><span><span>y2</span>−9</span><span>−−−−−</span>√</span>dy</span><span>=4π<span>∫43</span>y<span><span><span>y2</span>−9</span><span>−−−−−</span>√</span>dy</span><span>=4π<span><span>[<span><span>13</span>(<span>y2</span>−9<span>)<span>32</span></span></span>]</span>43</span></span><span>=4π<span><span>[<span><span>13</span>(<span>y2</span>−9<span>)<span>32</span></span></span>]</span>43</span></span><span>=4π<span>[<span><span>7<span>7–√</span></span>3</span>]</span></span><span>=<span><span>28π<span>7–√</span></span>3</span></span></span></span></span>
Step-by-step explanation: Find The Radius Of A Quarter (Half Of The Diameter) Then Square The Radius, Or Multiply It By Itself. Then, Multiply The Squared Radius By Pi, Or 3.14, To Get The Area.