Question

Find the volume of the solid generated by revolving the region R bounded by $y = e^{-x}$, $y = 0$, $x = 0$ and $x = \ln 11$ about the x-axis. Set up the integral that gives the volume of the solid. $\int_0^{\ln 11} \boxed{\pi (e^{-x})^2} dx$ (Type exact answers.)

          Find the volume of the solid generated by revolving the region R bounded by $y = e^{-x}$, $y = 0$, $x = 0$ and $x = \ln 11$ about the x-axis.
Set up the integral that gives the volume of the solid.
$\int_0^{\ln 11} \boxed{\pi (e^{-x})^2} dx$
(Type exact answers.)

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Find the volume of the solid generated by revolving the region R bounded by y = e^-x, y = 0, x = 0 and x = ln 11 about the x-axis.
Set up the integral that gives the volume of the solid.
∫0^ln 11π (e^-x)^2 dx
(Type exact answers.)

Added by Shelly E.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Steven Clarke

12/01/2022

Step 1

Step 1: The volume of the solid generated by revolving the region R bounded by y = e x, y = 0, x = 0 and x = In 11 about the x-axis is given by the integral: $$ \int_{0}^{\ln 11} \pi (e^{-x})^2 dx $$ Show more…

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Find the volume of the solid generated by revolving the region R bounded by y = e x, y = 0, x = 0 and x = In 11 about the x-axis. Set up the integral that gives the volume of the solid. $$ \int_{0}^{\ln 11} $$ dx (Type exact answers.) R y = e x In 11

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Transcript

00:01 Okay, here i want to work out the volume and the green area here is rotated about the x -axis.

00:09 Now here we use the formula for rotation, which is this.

00:14 The volume is integral a to b, pi, y squared, dx.

00:23 That's rotation about the x axis from a to b.

00:28 So here we're doing two parts.

00:30 What i do first is i work out if i rotate this area under the curve y equals e power x over 2 about the x axis between these two limits here.

00:49 And then what i do is i will take away this volume here under the curve y equals e to minus x over 2.

00:59 And that's the answer.

01:04 So what we have then is this.

01:09 The required volume is the integral between ln6 and ln11, pi, the top one, this one squared.

01:23 So is e to the y over 2, 4 squared, minus the bottom one.

01:31 So integral, ln6, ln11, pi, e minus x over 2.

01:42 I'm not y there, x of course, and here squared.

01:55 How many dx here? i need dx here.

01:59 Well, this i can combine into a single integral.

02:03 It's going to be integral ln6 to ln11.

02:08 Take the pi outside.

02:11 I work at this, e to the x over 2 are squared.

02:15 Third law of exponents.

02:17 We multiply this by 2...

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