Answer to Question #221881 in Calculus for Anuj

Question #221881
Consider the surface S = n (x, y, z) | z = p x 2 + y 2 and 1 ≤ z ≤ 3 o .(a) Sketch the surface S in R 3 . Also show its XY-projection on your sketch. (2) (b) Evaluate the area of S, using a surface integral
1
Expert's answer
2021-11-02T18:45:20-0400

Answer:-

Part a

"z=x^2+y^2\\\\\n1\u2264z\u22643\\\\\n1\u2264x^2+y^2\u22643\\\\\n1^2\u2264x^2+y^2\u2264(\\sqrt3)^2\\\\"



The XY-projection the sketch is the shaded region

Part b

"z=x^2+y^2\\\\\nzx=2x\\\\\nzy=2y\\\\"

Change in polar coordinate

"x=r cos \\theta; y=r sin \\theta\\\\\nx^2+y^2=r^2; dA=rdrd \\theta\\\\\n1<r< \\sqrt{3}; 0< \\theta < 2\\pi\\\\\n\\sqrt{1+zx^2+zy^2}= \\sqrt{1+4x^2+4y^2}= \\sqrt{1+4r^2}"

Surface area

"=\\int \\int _R \\sqrt{1+zx^2+zy^2}\\\\\n =\\int_0^{2 \\pi} \\int_1^{\\sqrt{3}} \\sqrt{1+4r^2} rdrd \\theta\\\\\n=\\int_0^{2 \\pi} d \\theta \\int_1^{\\sqrt{3}} \\sqrt{1+4r^2} rdr\\\\\nLet \\space 1+4r^2 = t^2\\\\\n=(2 \\pi -0) \\int _{\\sqrt{5}}^{\\sqrt{3}} t * \\frac{t dt}{4} \\\\\n= \\frac{2 \\pi}{4}[\\frac{t^3}{3}]_{\\sqrt5}^{\\sqrt3}\\\\\n=\\frac{\\pi}{6}[13 \\sqrt{13}-5 \\sqrt5]"



The surface S in R3

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