Question

The temperature in degrees Celsius on the surface of a metal plate is T(x, y) = 19 − 4x2 − y2 where x and y are measured in centimeters. Estimate the average temperature when x varies between 0 and 2 centimeters and y varies between 0 and 4 centimeters. °C

Answer 1

Answer:

Average value of temperature will be $8.335^{\circ}C$

Step-by-step explanation:

We have given the temperature in degree Celsius $T(x,y)=19-4x^2-y^2$

It is given that x varies between 0 to 2

So $0\leq x\leq 2$

And y varies between 0 and 4

So $0\leq y\leq 4$

So area between the region A = 4×2 = 8$cm^2$

Now average temperature is given by

$Average\ value=\frac{1}{A}\int \int T(x,y)dA$

$Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy$

$=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy$

As limit of x is 0 to 2

$=\frac{1}{8}\int  (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy$

=$=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})$

As limit of y is 0 to 4

So $Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C$