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Rework Example 6 assuming that the water tank is in the shape of an upper hemisphere of radius 20 feet. (See Problem 21 for the volume of a spherical segment.)...

Question

Rework Example 6 assuming that the water tank is in the shape of an upper hemisphere of radius 20 feet. (See Problem 21 for the volume of a spherical segment.)

Rework Example 6 assuming that the water tank is in the shape of an upper hemisphere of radius 20 feet. (See Problem 21 for the volume of a spherical segment.)



Answers

Using a Sphere A sphere of radius $r$ is cut by a plane $h$
units above the equator, where $h<r .$ Find the volume of the
solid (spherical segment) above the plane.

Were asked to find the volume of the solid that lies within both the cylinder X squared plus y squared equals one, and this feared X squared plus y squared plus Z squared equals four. So in cylindrical coordinates, region of Integration E. This is the solid region that is within the cylinder R squared equals one or R equals one and bounded above and below by the sphere. R Squared plus Z squared equals four, and therefore he is the set of triples. Our Fate. A Z such that data lies between zero and two pi. Our lives between zero and one and Z Rearranging lies between the negative square root to four minus R squared and the positive square root of four minus r squared. And so the volume of this solid E is the triple integral of one over E, which is the integral from 0 to 2 pi integral from 0 to 1, integral from Route four minus R squared. So are negative. Route four minus R squared, deposited root for minus R squared of the differential, which is are D C D r d theta segment, a derivative with respect to Z and evaluating. We get integral from 0 to 2 pi integral from 01 of to our times the square root of four minus R squared de are de Seda using clubbiness theory we can right. This is a product of integral integral from 0 to 2 pi di fada times integral from 01 to our times, The square two for minus R squared you are and performing a use substitution in our heads and evaluating taking into derivatives. First we get two pi times and then here we have negative 1/3 halves, which is negative two thirds times the inside function for minus R squared the three halves evaluated from zero toe one. So the two and the one half from the exploding cancel out and we get evaluating four thirds pi times eight minus three to the three hats.

Were asked to find the volume of the solid that is enclosed by the cone. Z equals the square root of four X squared plus y squared in the sphere X squared plus y squared plus C squared equals two. So if you visualize this region in your head, reading E is bounded below by the cone. Z equals square root of R squared, which is simply Z equals R and is bounded above by the sphere C squared plus R squared equals two. And since really concerned about the top part of this fear, this is really Z equals the positive square root of two minus r squared and therefore well, we also know that the cone and sphere intersect when they're Z values are equal. So when we have that two times R squared is equal to two according to our 1st and 2nd equations. So in other words, are is you put a one. And so the region E is the set of triples R theta Z such that well, you know our is going fatal lies between zero and two pi. Our lives between zero and the point of Intersection one and Z lies well has to lie between the cone or and below the sphere square root of two minus R squared and therefore the volume of this region is the triple integral of one over E. And this is integral from 0 to 2 pi integral from 0 to 1, integral from our to the square root of two minus R squared of the differential or dizzy D r d theta Taking the anti derivative with respect to Z, we get integral from 0 to 2 pi integral from 01 and then our times e from Z equals R to Z equals the square root of two minus R squared DRD theta and evaluating. We get integral from 0 to 2 pi integral from zero toe one Then this is are times that squared to minus R squared, minus R squared DRB theta and taking the anti derivative first. Actually, that's useful. Beanies. The're, um right. This is a proactive, integral. So you have been a girl from 0 to 2. Pi do data times integral from 01 are Times Square to minus R squared minus R squared D r now taking anti derivatives. We get two pi times and this is doing a use substitution in her head. Negative One half times one over three halves. So negative one third times are inter function to minus R squared to the three halves. Minus one third are a cubed from 0 to 1. Evaluating we get two pi times negative one third times and then we have one minus 10 Sorry, one plus one. My mistake. And then minus zero. Yeah, well, actually minus two to the three halves minus zero which simplifies to negative two thirds pi times. This is two minus two route to distributing the negative in factoring This is the same as four thirds pi times route to minus one.

Were asked to find the volume of this solid that lies between the parable Lloyd Z equals X Squared plus y squared and the Spear X squared plus y squared plus Z squared equals two. So thinking in terms of cylindrical coordinates, we know this region E is bounded below by the parable. Lloyd with Equation Z equals R squared and is bounded above by the sphere with equation R squared plus C squared equals two. Or since we're only concerned about the top part of this fear, Z equals the positive square roots. Tu minus r squared. Now these two surfaces the parable Lloyd and Spear intersect when we have that. Both sequel R Squared and R Squared plus C squared equals two. So in other words, when R Squared plus part of the fourth equals two and factoring we get r squared. So are the fourth plus R squared minus two, which is R squared minus two times R squared. Sorry I squared, plus two times R squared minus one equals zero. And solving we get either r squared equals negative two, which is impossible or are scrape equals one. So it follows the r squared equals one and therefore the R is equal to one since our must be positive. And so our region e is the set of triples our fate a Z such that fatal lies between zero in two pi are lies between zero and that curve of intersection at one and Z lies between the parable Lloyd R Squared and the cone square root of two minus R squared certain outcome the sphere. And so the volume of this region e is the triple integral of one over easy. This is integral from zero to pi integral from 01 integral from r squared to the square root of two minus r squared of the differential are dizzy DRD data taking the anti derivative With respect Dizzy, we get integral from 0 to 2 pi integral from 01 of our Z from Z equals r squared to Z equals the squared of two minus r squared DRG theta evaluating we get the integral from 0 to 2 pi integral from 01 of our times The square root of two minus R squared minus R cubed d r d theta using Fabini steer American right. This is a product of integral so we have the integral from 0 to 2 pi d theta times the interval from 01 of our times that squared of two minus r squared minus are cute D R and taking anti derivatives and evaluating. We get two pi times and here you'll make a U substitution your head so we have negative one half times 1/3 halves is negative. One third times the inter function to minus R squared to the three hands minus 1/4 Art of the fourth from 01 and evaluating this is two pi times negative. One third minus 1/4 plus one third times, two to the three halves minus zero and this simplifies to two pi times negative 7/12 plus two thirds route to and multiplying out the two we get negative 7/6 plus four thirds route to times pi.

Were asked to you spiritually coordinates to find the volume of the solid that lies within the sphere X squared plus y squared policies critical is for above the X Y plane and below the cone Z equals the square root of X squared plus y squared. So spherical coordinates sphere X squared plus y squared plus Z squared equals four. This is equivalent to the equation. Row equals the square root of four or two. In the cone, Z equals the positive square root of X squared plus y squared. This is represented in spherical coordinates as Z equals Sorry, Nazi Phi equals pi over four. Yeah, and so the solid is given by instead of triples. Row data fight such that road lies between zero and two. The radius of the sphere data lies between zero two pi and five will lie between where we have to be above the X Y plane but below the cone. So fi lies between pi over four cone and the X Y plane, which is five, equals pi over two. And therefore the volume of this solid is the integral from zero. Sorry. Pirate 42 pi over two. Integral from 0 to 2 pi integral from 0 to 2 of the differential row squared sine phi. Do you row de theater defy taking the anti derivative with respect Roe Fi and data first, Actually, we'll use diabetes. The're, um right. This is a product of inch girls. So we have integral from pirate 42 pi over two of sign five defy times the integral from 0 to 2 pi d theta times had to grow from 0 to 2 of row squared zero taking into derivatives. This is negative cosine phi from PIRA four However, to times data from 0 to 2 pi times one third row cubed from 0 to 2 Evaluating this is to over to times two pi times eight thirds Simplifying this is eight times route to over three times pi


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