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Consider the solid obtained by rotating the region bounded by the given curves about the y-axis. $y = \ln 4x$, $y = 2$, $y = 3$, $x = 0$ Find the volume $V$ of this solid. Part 1 of 6 We will use horizontal strips. Such a strip has a vertical span of $\frac{\pi}{32}\left(e^{6}-e^{4}\right)$ $\times$ $\Delta y$ $x$ $y$ $x=0$ $y_{1}$ $y_{2}$ $y_{3}$ $0$ $x$ $y$ $0$ $x$ $y_{1}=3$ $y_{2}=2$ $y_{3}=\ln 4x$ Part 2 of 6 We must re-write $y = \ln 4x$ as $x = \frac{e^{(y)}}{4}$ $\frac{1}{4}e^{y}$ Part 3 of 6 When the shaded region is rotated around the y-axis, a horizontal strip of vertical span $\Delta y$ creates a disk. The radius of this disk is $r = \frac{e^{y}}{4}$ $\frac{1}{4}e^{y}$ Part 4 of 6 The cross-sectional area of a disk of radius $r$ is $A = \pi r^{2}$ $\pi r^{2}$ Since our disk has radius $r = \frac{1}{4}e^{y}$, then the cross-sectional area is $A = \frac{\pi}{16}e^{2y}$ $\pi\left(\frac{1}{4}e^{y}\right)^{2}$ Part 5 of 6 Now we can say that the volume of the solid created by rotating the shaded area around the y-axis is $V = \int_{c}^{d} A(y) d y = \int_{2}^{3} \pi\left(\frac{1}{4} e^{y}\right)^{2} d y$.

          Consider the solid obtained by rotating the region bounded by the given curves about the y-axis.
$y = \ln 4x$, $y = 2$, $y = 3$, $x = 0$
Find the volume $V$ of this solid.

Part 1 of 6
We will use horizontal strips.

Such a strip has a vertical span of $\frac{\pi}{32}\left(e^{6}-e^{4}\right)$ $\times$ $\Delta y$

$x$
$y$
$x=0$
$y_{1}$
$y_{2}$
$y_{3}$
$0$
$x$
$y$
$0$
$x$
$y_{1}=3$
$y_{2}=2$
$y_{3}=\ln 4x$

Part 2 of 6
We must re-write $y = \ln 4x$ as $x = \frac{e^{(y)}}{4}$ $\frac{1}{4}e^{y}$

Part 3 of 6
When the shaded region is rotated around the y-axis, a horizontal strip of vertical span $\Delta y$ creates a disk.

The radius of this disk is $r = \frac{e^{y}}{4}$ $\frac{1}{4}e^{y}$

Part 4 of 6
The cross-sectional area of a disk of radius $r$ is $A = \pi r^{2}$ $\pi r^{2}$

Since our disk has radius $r = \frac{1}{4}e^{y}$, then the cross-sectional area is

$A = \frac{\pi}{16}e^{2y}$ $\pi\left(\frac{1}{4}e^{y}\right)^{2}$

Part 5 of 6
Now we can say that the volume of the solid created by rotating the shaded area around the y-axis is $V = \int_{c}^{d} A(y) d y = \int_{2}^{3} \pi\left(\frac{1}{4} e^{y}\right)^{2} d y$.

consider the solid obtained by rotating the region bounded by the given curves about the y axis y ln 4x y 2 y 3 x 0 find the volume v of this solid part 1 of 6 we will use horizontal strips 19971

Added by Samantha H.

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