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Part 1 of 8 By re-ordering the sphere equation, we have z- 100 - - y, We can then parameterize this surface in rectangular coordinates as 2-V100 - 2 V100-- Part 2 of 8 Next, we need to find a restriction for u and v to describe only the part of the sphere that lies above the cone. We start by examining how the sphere and the cone intersect. Since both surfaces are symmetric around the z-axis, the sphere x++ - 100 intersects the cone z v in a circle. To find the equation of this cirele, we begin by substituting z V+y for z in the sphere equation, to get * y• ( y - 2., 2- 100. This can be further simplified to +- 50 P so Part 3 of8 The equation x+ y- 50 describes a cylinder, and the intersection circle is above the xy-plane and en this cylinder. The z-value of points on this circle can be found by solving z and x+- 50 simultaneously to find z, thereby obtaining z-V50 Part 4 of Since we want the portion of the sphere which is above this circle, we return to the parametric representation zV100-u - and limit z to be at least 50. We, therefore, have V 100-u - 2 vs0, which means u + s 5o. Part 5 of An alternative approach involves using spherical coordinates o, e, e), In spherical coordinates, the sphere has the equation p- 10 10 Part 6 of 8 Therefore, we can use and p as parameters and use the equations that allow us to convert from spherical to rectangular coordinates. We would then have *-10 sin cos e Subrmit Ske (you cancot.come back

Part 1 of 8 By re-ordering the sphere equation, we have z- 100 - - y, We can then parameterize this surface in rectangular coordinates as 2-V100 - 2 V100-- Part 2 of 8 Next, we need to find a restriction for u and v to describe only the part of the sphere that lies above the cone. We start by examining how the sphere and the cone intersect. Since both surfaces are symmetric around the z-axis, the sphere x++ - 100 intersects the cone z v in a circle. To find the equation of this cirele, we begin by substituting z V+y for z in the sphere equation, to get * y• ( y - 2., 2- 100. This can be further simplified to +- 50 P so Part 3 of8 The equation x+ y- 50 describes a cylinder, and the intersection circle is above the xy-plane and en this cylinder. The z-value of points on this circle can be found by solving z and x+- 50 simultaneously to find z, thereby obtaining z-V50 Part 4 of Since we want the portion of the sphere which is above this circle, we return to the parametric representation zV100-u - and limit z to be at least 50. We, therefore, have V 100-u - 2 vs0, which means u + s 5o. Part 5 of An alternative approach involves using spherical coordinates o, e, e), In spherical coordinates, the sphere has the equation p- 10 10 Part 6 of 8 Therefore, we can use and p as parameters and use the equations that allow us to convert from spherical to rectangular coordinates. We would then have *-10 sin cos e Subrmit Ske (you cancot.come back

Algebra and Trigonometry (MindTap Course List)
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter12: Conic Sections
Section12.CR: Chapter Review
Problem 6CC
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Part 1 of 8 By re-ordering the sphere equation, we have z? = 100 – x2 - y2. We can then parameterize this surface in rectangular coordinates as X = u y = v V100 – u? – v² Z = 100 – u? v2 Part 2 of 8 Next, we need to find a restriction for u and v to describe only the part of the sphere that lies above the cone. We start by examining how the sphere and the cone intersect. Since both surfaces are symmetric around the z-axis, the sphere x2 + y2 + z? = 100 intersects the cone z = x² + y2 in a circle. To find the equation of this circle, we begin by substituting z = V x2 + y? for z in the sphere equation, to get x2 + y? + (Vx2 + y?)² = 2x² + y? = 100. This can be further simplified to x² + y2 = = 50 50 Part 3 of 8 The equation x² + y2 = 50 describes a cylinder, and the intersection circle is above the xy-plane and on this cylinder. The z-value of points on this circle can be found by solving z = Vx2 + y? and x2 + y2 = 50 simultaneously to find z, thereby obtaining z = 50 V50 Part 4 of 8 Since we want the portion of the sphere which is above this circle, we return to the parametric representation z = 100 – u? - v2 and limit z to be at least V50. We, therefore, have V 100-u? -v2 = z > V50, which means u2 + v2 50. Part 5 of 8 An alternative approach involves using spherical coordinates (p, 0, 4). In spherical coordinates, the sphere has the equation p = 10 10 Part 6 of 8 Therefore, we can use e and o as parameters and use the equations that allow us to convert from spherical to rectangular coordinates. We would then have X = 10 sin o cos e y = z = Submit Skip (you cannot come back)
Transcribed Image Text:Part 1 of 8 By re-ordering the sphere equation, we have z? = 100 – x2 - y2. We can then parameterize this surface in rectangular coordinates as X = u y = v V100 – u? – v² Z = 100 – u? v2 Part 2 of 8 Next, we need to find a restriction for u and v to describe only the part of the sphere that lies above the cone. We start by examining how the sphere and the cone intersect. Since both surfaces are symmetric around the z-axis, the sphere x2 + y2 + z? = 100 intersects the cone z = x² + y2 in a circle. To find the equation of this circle, we begin by substituting z = V x2 + y? for z in the sphere equation, to get x2 + y? + (Vx2 + y?)² = 2x² + y? = 100. This can be further simplified to x² + y2 = = 50 50 Part 3 of 8 The equation x² + y2 = 50 describes a cylinder, and the intersection circle is above the xy-plane and on this cylinder. The z-value of points on this circle can be found by solving z = Vx2 + y? and x2 + y2 = 50 simultaneously to find z, thereby obtaining z = 50 V50 Part 4 of 8 Since we want the portion of the sphere which is above this circle, we return to the parametric representation z = 100 – u? - v2 and limit z to be at least V50. We, therefore, have V 100-u? -v2 = z > V50, which means u2 + v2 50. Part 5 of 8 An alternative approach involves using spherical coordinates (p, 0, 4). In spherical coordinates, the sphere has the equation p = 10 10 Part 6 of 8 Therefore, we can use e and o as parameters and use the equations that allow us to convert from spherical to rectangular coordinates. We would then have X = 10 sin o cos e y = z = Submit Skip (you cannot come back)
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