Answer to Question #253100 in Calculus for Anuj

Question

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

Answer 1

Part a

$z=x^2+y^2\\ 1≤z≤3\\ 1≤x^2+y^2≤3\\ 1^2≤x^2+y^2≤(\sqrt3)^2\\$

The XY-projection the sketch is the shaded region

Part b

$z=x^2+y^2\\ zx=2x\\ zy=2y\\$

Change in polar coordinate

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

Surface area

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

The surface S in R³