Question

(1 point) The soot produced by a garbage incinerator spreads out in a circular pattern. The depth, H(r), in millimeters, of the soot deposited each month at a distance r kilometers from the incinerator is given by H(r)=0.114e−1.8r. (a) Write a definite integral (with independent variable r) giving the total volume of soot deposited within 5 kilometers of the incinerator each month. volume = ∫50 (include units) (b) Evaluate the integral you found in part (a) to find the volume of soot.

Answer 1

Answer:

a) $V(r) = \int\limits^5_00.114e^{-1.8r} \, dr$

b) $V = 0.0629$

Step-by-step explanation:

We have that the depth, in millimeters, of the soot deposited each month at a distance r kilometers from the incinerator is given by:

$H(r) = 0.114e^{-1.8r}$

(a) Write a definite integral (with independent variable r) giving the total volume of soot deposited within 5 kilometers of the incinerator each month.

This is the integral of H(r) with r varying from 0 to 5. So:

$V(r) = \int\limits^5_0H(r) \, dr$

$V(r) = \int\limits^5_00.114e^{-1.8r} \, dr$

(b) Evaluate the integral you found in part (a) to find the volume of soot.

$V(r) = \int\limits^5_00.114e^{-1.8r} \, dr$

$V(r) = 0.114\int\limits^5_0e^{-1.8r} \, dr$

$V(r) = -\frac{0.114}{1.8}e^{-1.8r}, 0\leq r\leq5$

$V(r) = -0.063e^{-1.8r}, 0\leq r\leq5$

$V = V(5) - V(0)$

$V = -0.00000777481 - (-0.063)$

$V = 0.0629$