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Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. y = x² + 8 y = -x² + 2x + 12 x = 0 X = 3 Step 1 To find the point(s) of intersection of the curves y = x² + 8 and y = -x² + 2x + 12, equate both equations and solve. (x - x² - 2x - x² - x - )(x + x² + 8 = -x² + 2x + 12 = 0 = 0 ) = 0 X = , X =

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. y = x² + 8 y = -x² + 2x + 12 x = 0 X = 3 Step 1 To find the point(s) of intersection of the curves y = x² + 8 and y = -x² + 2x + 12, equate both equations and solve. (x - x² - 2x - x² - x - )(x + x² + 8 = -x² + 2x + 12 = 0 = 0 ) = 0 X = , X =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.3: Hyperbolas
Problem 37E
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### Finding the Volume of a Solid Generated by Revolving a Region Around the x-axis To find the volume of the solid generated by revolving the region bounded by the graphs of the equations around the x-axis, follow these steps: Given equations: \[ y = x^2 + 8 \] \[ y = -x^2 + 2x + 12 \] \[ x = 0 \] \[ x = 3 \] #### Step 1: **To find the point(s) of intersection of the curves \( y = x^2 + 8 \) and \( y = -x^2 + 2x + 12 \).** We equate the two equations and solve for x. \[ x^2 + 8 = -x^2 + 2x + 12 \] Rearrange all terms to one side of the equation for standard quadratic form: \[ x^2 + 8 + x^2 - 2x - 12 = 0 \] Simplify: \[ 2x^2 - 2x - 4 = 0 \] Divide throughout by 2 for simplicity: \[ x^2 - x - 2 = 0 \] Factorize the quadratic equation: \[ (x - 2)(x + 1) = 0 \] Solve for x: \[ x = 2 \] \[ x = -1 \] Therefore, the points of intersection are \( x = 2 \) and \( x = -1 \). These points will be crucial in determining the limits of integration for setting up the volume integral in the subsequent steps.
Transcribed Image Text:### Finding the Volume of a Solid Generated by Revolving a Region Around the x-axis To find the volume of the solid generated by revolving the region bounded by the graphs of the equations around the x-axis, follow these steps: Given equations: \[ y = x^2 + 8 \] \[ y = -x^2 + 2x + 12 \] \[ x = 0 \] \[ x = 3 \] #### Step 1: **To find the point(s) of intersection of the curves \( y = x^2 + 8 \) and \( y = -x^2 + 2x + 12 \).** We equate the two equations and solve for x. \[ x^2 + 8 = -x^2 + 2x + 12 \] Rearrange all terms to one side of the equation for standard quadratic form: \[ x^2 + 8 + x^2 - 2x - 12 = 0 \] Simplify: \[ 2x^2 - 2x - 4 = 0 \] Divide throughout by 2 for simplicity: \[ x^2 - x - 2 = 0 \] Factorize the quadratic equation: \[ (x - 2)(x + 1) = 0 \] Solve for x: \[ x = 2 \] \[ x = -1 \] Therefore, the points of intersection are \( x = 2 \) and \( x = -1 \). These points will be crucial in determining the limits of integration for setting up the volume integral in the subsequent steps.
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