<Text-field style="Heading 1" layout="Heading 1">Introduction</Text-field>

LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0Yn

Let us define a bowl as the surface of revolution for a curve LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEieUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIj1GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUS90aGlja21hdGhzcGFjZUYnLyUncnNwYWNlR0ZPLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnLUYsNiVRImZGJ0YvRjItSShtZmVuY2VkR0YkNiQtRiM2Iy1GLDYlUSJ4RidGL0YyRjk=.

f:= x->3-sqrt(9-x^2);

plot(f(x),x=0..3,scaling=constrained);

We're going to rotate the curve around the y-axis. Here is how we can plot the surface. (You'll learn more about this type of plot if you take Math 243.)

plot3d([x*cos(t),x*sin(t),f(x)],x=0..3,t=0..2*Pi,scaling=constrained);

The picture suggests that the 'bowl' is a hemisphere, which in fact it is.

Let's compute the volume within the bowl. We will approach this by integrating horizontal disks, using y as the integration variable. Assuming for now that each disk has a radius LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEickYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUYsNiVRInlGJ0YvRjIvRjNRJ25vcm1hbEYn, this means that

vol:= Int( Pi*r(y)^2, y=0..3 );

The radius LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEickYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUYsNiVRInlGJ0YvRjIvRjNRJ25vcm1hbEYn of each disk equals x when 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. In other words, 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. We can try to find the inverse function by using solve.

solve(y=f(x),x);

Maple found two choices for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw==, one positive and one negative. Since we have said the domain of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw== is between 0 and 3, we want the positive choice. For me, that is the first result, so I will use %[1] to extract the first expression. However, the solve results can be returned in different orders, so check whether you have to change the next line.

r:=unapply(%[1],y);

Notice that domain of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEickYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw== is the range of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw==, that is, the interval [0,3].

plot( [f(x),r(x),x], x=0..3, color=[red,blue,black], scaling=constrained );

The inverse functions have the required mirror symmetry.

Now we are ready to compute the volume. Since we have defined LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEickYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYjLUYsNiVRInlGJ0YvRjIvRjNRJ25vcm1hbEYn, the volume integral is better defined too:

vol;

value(%);

Of course, this is a hemisphere, so we can check that this agrees with LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkmbWZyYWNHRiQ2KC1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictRiM2Iy1GLzYkUSIzRidGMi8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGPy8lKWJldmVsbGVkR1EmZmFsc2VGJy1JI21pR0YkNiVRJyYjOTYwO0YnLyUnaXRhbGljR0ZERjItSSVtc3VwR0YkNiVGN0Y1LyUxc3VwZXJzY3JpcHRzaGlmdEdRIjBGJw==.

3^3*2/3*Pi;

It's more interesting to be able to compute the volume of water partially filling the bowl up to arbitrary height. In fact we can describe the volume as a function of the water height LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiaEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIi5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGTy8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJw==

V:= h -> int( Pi*r(y)^2, y=0..h );

In a function definition, Maple leaves the integral unevaluated. But we can ask for the value of the function, and get

V(h);

plot(V(h),h=0..3);

Finally, we could calibrate a dipstick that would tell us at what heights the bowl is 1/4, 1/2, and 3/4 full (that's 3/4, 1/2, and 1/4 empty for you pessimists). This is a case where the floating point fsolve serves us better than solve does.

fsolve( V(h)=V(3)/4, h );

fsolve( V(h)=V(3)/2, h );

fsolve( V(h)=3*V(3)/4, h );

For my run, the physically relevant answer is the middle one in each case.