Loading [MathJax]/jax/output/HTML-CSS/jax.js
AMPS | THARC | KE8QZC | SFW | TSW | WW
ORCID iD icon

Back to the class

Section 7.2 #24: Find the volume of the solid generated by revolving the region bounded by y=x4x2 and y=0 about the x-axis.
Solution: Sketch the region and washers:
Note that we will use two integrals to compute the area of the solid of revolution. This is for the same reason we would have to use two integrals to compute the area of the region: the "top" function and the "bottom" function change after the intersection point. Therefore compute Volume=π02[x4x2]2dx+π20[x4x2]2dx=π024x2x4dx+π204x2x4dx=π[43x315x5|02+π[43x315x5|20=π{0[323+325]}+π{[323325]0}=π[643645]=128π15.

Section 7.2 #28: Find the volume of the solid generated by revolving the region bounded by y=ex4,y=0,x=0, and x=6 about the x-axis.
Solution: Sketch the region:
Now sketch the revolved solid and a washer:
The radius of the washer is R(x)=ex4, so we can now compute the volume. Compute Volume=π60(ex4)2dx=π60ex2dxu=x2=2π30eudu=2π[eu|30=2π(e31).

Section 7.2 #33: Find the volume of the solid generated by revolving the region bounded by the graphs of y=sin(x),y=0,x=0, and x=π about the x-axis.
Solution: First sketch the region:
Now sketched the revolved region and a washer:
Recall the trigonometric identity sin2(x)=1cos(2x)2. Now find the volume: compute Volume=ππ0sin2(x)dx=ππ01cos(x)2dx=π2π01dxπ2π0cos(x)dx=π2[x|π0π2[sin(x)|π0=π2(π0)π2(00)=π22.

Section 7.3 #8: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y=9x2 and y=0 about the y-axis.
Solution: Draw the region:
and draw the shell:
Note that when this region rotates around the x-axis, the "right half" will intersect the path of the "left half". To avoid overcounting the volume, we will simply rotate the right half of the region instead. Now compute Volume=2π30x(9x2)dx=2π[92x214x4|30=2π(812814)=162π4=81π2.

Section 7.3 #14: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y={sin(x)x,x>01,x=0},y=0,x=0, and x=π about the y-axis.
Solution: First draw the region
and draw a shell
Therefore compute Volume=2ππ0x(sin(x)x)dx=2ππ0sin(x)dx=2π[cos(x)|π0=2π[(1)(1)]=4π.

Section 7.3 #20: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y=4x2, x=0, and y=4 about the x-axis.
Solution: First note that the intersection point of y=4 and y=4x2 is at x=1 with height y=4. Now draw the region
and draw a shell
Therefore calculate Volume=2π4012y32dy=2π[15y52|40=64π5. Section 7.3 #24: Use the shell method to find the volume of the solid generated by revolving the region bounded by y=x,y=0, and y=4 about the line x=6.
Solution: First draw the region
and draw the shell
Now calculate Volume=2π40x(6x)dx=12π40x12dx2π40x32dx=12π[23x32|402π[25x52|40=24π3[4320]4π5[4520]=64π128π5=320π5128π5=192π5.

Section 7.3 #25: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y=x2 and y=4xx2 about the line x=4.
Solution: First draw the region
and the shell
Now calculate Volume=2π20(4x)((4xx2)x2)dx=2π20(4x)(4x2x2)dx=2π202x312x2+16xdx=2π[24x4123x3+162x2|20=2π[x424x3+8x2|20=2π[(832+32)0]=16π.