Question

Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral. The part of the surface that lies above the disk x2 + y2 ≤ 81

Answer 1

Answer:

A(s) = 255.8857

Step-by-step explanation:

Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral. The part of the surface z = e^-x^2-y^2 that lies above the disk x2 + y2 ≤ 81.

Given that:

$Z = e^{-x^2-y^2}$

By applying rule; the partial derivatives with respect to x and y

$\dfrac{\partial z }{\partial x}= -2xe^{-x^2-y^2}$

$\dfrac{\partial z }{\partial y}= -2ye^{-x^2-y^2}$

The integral over the general region D with respect to x and y is :

$A(s) = \int \int _D \sqrt{1+(\dfrac{\partial z}{\partial x} )^2 +(\dfrac{\partial z}{\partial y} )^2 }\ dA$

$A(s) = \int \int _D \sqrt{1+(-2xe^{-x^2-y^2})^2 +(-2ye^{-x^2-y^2})^2 } \ dA$

$A(s) = \int \int _D \sqrt{1+4x^2({e^{-x^2-y^2})^2 +4y^2({e^{-x^2-y^2}})^2 }} \ dA$

$A(s) = \int \int _D \sqrt{1+(4x^2+4y^2)({e^{-x^2-y^2})^2 }} \ dA$

$A(s) = \int \int _D \sqrt{1+(4x^2+4y^2)e^{-2}({{x^2+y^2}) }} \ dA$

By relating the equation to cylindrical coordinates

$A(s) = \int \int_D \sqrt{1+4r^2 e^{-2r^2} }. rdA$

The bounds for integration for the circle within the cylinder $x^2+y^2 =81$ is r =9

$A(s) = \int \limits ^{2 \pi}_{0} \int \limits^9_0 r \sqrt{1+4r^2 e^{-2r^2} }. dr d\theta$

$A(s) = {2 \pi} \int \limits^9_0 r \sqrt{1+4r^2 e^{-2r^2} }\ dr$

Using integral calculator to estimate the  integral,we have:

A(s) = 255.8857