Answer to Question #221081 in Calculus for ndiser

Question #221081

(Section 13.3 and Chapter 14) Let D be the region in R 3 p that lies inside the cone z = x 2 + y 2 above the plane z = 1 and below the hemisphere z = p 4 − x 2 − y 2 . (a) Sketch the region D in R 3 .(b) Express the volume of D as a sum of triple integrals, using cylindrical coordinates.


1
Expert's answer
2021-10-13T01:01:08-0400

Answer:-




"V=V_1+V_2=\\int_0^{2\\pi}\\int_0^1\\int_1^{\\sqrt{4-x^2-y^2}}rdzdrd\\phi+ \\int_0^{2\\pi}\\int_1^{\\sqrt{2}}\\int_r^{\\sqrt{4-x^2-y^2}}rdzdrd\\phi"

Explanation on bounds of intergals

1) Intersection of cone "z=\\sqrt{x^2+y^2}" and plane z=1 is "1=\\sqrt{x^2+y^2}"

or "x^2+y^2=1" This is the circle with radius 1 and it is upper bound by r in V1 and lower bound with respect to r in V2.

2) Intersection of the cone "z^2=x^2+y^2" and sphere "x^2+y^2+z^2=4" is line "x^2+y^2=2,\\space z=\\sqrt{2}" . This is circle on height "\\sqrt{2}" with radius "r=\\sqrt{2}" and "\\sqrt{2}" is upper bond by r in V2.

3) For V2 z changes from plane z=1 to sphere "z=\\sqrt{4-x^2+y^2}"

4) For V1 z changes from cone "z=\\sqrt{x^2+y^2}=r" to upper bound in spere "z=\\sqrt{4-x^2-y^2}" .

5) All regions possess circular symmetry that is why "\\phi" changes from 0 to "2\\cdot \\pi"

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