QuestionFind the volume when the area between $y=x^{2}$ and $y=x^{3}$ is revolved around $x=1$ . Also, find the volume for $y=\sqrt{x}$ , $y=0$ , and $x=9$ revolved around $x=9$ .

Studdy Solution

STEP 1

Assumptions1. The region enclosed by the curves $y=x^{}$ and $y=x^{3}$ is revolved about the line $x=1$ . . The volume of the solid formed is calculated using the method of cylindrical shells.
3. The formula for the volume of a cylindrical shell is $V =\pi \int_{a}^{b} r(x)h(x)dx$ , where $r(x)$ is the radius and $h(x)$ is the height of the cylindrical shell.

STEP 2

First, we need to find the intersection points of the curves $y=x^{2}$ and $y=x^{}$ to determine the limits of integration. We do this by setting the two equations equal to each other and solving for $x$ .
$x^{2} = x^{}$

STEP 3

olving the equation gives us the intersection points.
$x^{2} - x^{3} =0$ $x^{2}(1 - x) =0$ So, $x =0$ and $x =1$ are the intersection points.

STEP 4

Now, we need to find the radius and height of the cylindrical shell. The radius is the distance from the line $x=1$ to the $x$ -coordinate of a point on the curve, which is $1 - x$ . The height is the difference between the $y$ -coordinates of the points on the curves, which is $x^{3} - x^{2}$ .

STEP 5

Substitute these values into the formula for the volume of a cylindrical shell and evaluate the integral.
$V =2\pi \int_{0}^{1} (1 - x)(x^{3} - x^{2})dx$

STEP 6

Expand the integrand and evaluate the integral.
$V =2\pi \int_{0}^{1} (x^{3} - x^{4} - x^{2} + x^{3})dx$ $V =2\pi \int_{0}^{1} (2x^{3} - x^{4} - x^{2})dx$

STEP 7

Evaluate the integral.
$V =2\pi [ \frac{1}{2}x^{4} - \frac{1}{5}x^{5} - \frac{1}{3}x^{3}]_{0}^{1}$ $V =2\pi [ \frac{1}{2} - \frac{1}{5} - \frac{1}{3}]$

STEP 8

implify the expression to get the volume of the solid.
$V =2\pi [ \frac{10 -4 -6}{30}]$ $V =2\pi [0]$ $V =0$ So, the volume of the solid that results when the region enclosed by $y=x^{2}$ and $y=x^{3}$ is revolved about the line $x=1$ is0.

STEP 9

Assumptions. The region enclosed by the curves $y=\sqrt{x}, y=$ , and $x=9$ is revolved about the line $x=9$ .
2. The volume of the solid formed is calculated using the method of cylindrical shells.
3. The formula for the volume of a cylindrical shell is $V =2\pi \int_{a}^{b} r(x)h(x)dx$ , where $r(x)$ is the radius and $h(x)$ is the height of the cylindrical shell.

STEP 10

The radius is the distance from the line $x=9$ to the $x$ -coordinate of a point on the curve, which is $9 - x$ . The height is the $y$ -coordinate of a point on the curve, which is $\sqrt{x}$ .

STEP 11

Substitute these values into the formula for the volume of a cylindrical shell and evaluate the integral.
$V =\pi \int_{0}^{9} (9 - x)\sqrt{x}dx$

STEP 12

Use the substitution $u =9 - x$ , $du = -dx$ , $x =9 - u$ to simplify the integral.
$V =2\pi \int_{0}^{9} u\sqrt{9 - u}(-du)$

STEP 13

Evaluate the integral.
$V =2\pi [ -\frac{2}{5}(9 - u)^{5/2}]_{0}^{9}$ $V =2\pi [ -\frac{2}{5}(0) + \frac{2}{5}(9)^{5/2}]$

STEP 14

implify the expression to get the volume of the solid.
$V =2\pi [ \frac{2}{}(9)^{/2}]$ $V = \frac{4\pi}{}(243)$ $V = \frac{972\pi}{}$ So, the volume of the solid that results when the region enclosed by $y=\sqrt{x}, y=0$ , and $x=9$ is revolved about the line $x=9$ is $\frac{972\pi}{}$ .

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