3 volume via integration problems that are giving me trouble

  • Thread starter JasonJo
  • Start date
  • #1
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2
my teacher never actually went over this, but i tried doing them anyway. there were 15 problems originally assigned, and i wanted to get them all done on my own before class on tuesday, im hoping you guys can help out

12) You want to make a pyramid with height = 3260 and square base with side s=1780

a) find the integrand
b) find the volume

i know the volume of a pyramid is Bh/3, but i dunno how to get the area of the delta x and then sum infinite delta x's together

14) Find the volume of the solid formed by rotating the region enclosed by:
y= e^(4x) + 1
y = 0
x = 0
x = 0.5

15) find for the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^3
y=25x

do i go about finding the volume of the entire solid if i were to rotate it around and then cut it in half?

thanks a ton guys, i usually get great hw from this forum
 

Answers and Replies

  • #2
StatusX
Homework Helper
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12) If you can get the length of the sides of the pyramid at a given z-value, you can get the area of the cross-sections in terms of z. Integrate this over z.

14) Just integrate the function from 0 to 0.5.

15) I don't understand you're question about cutting it in half, but just use the washer method, integrating [itex]\pi((25x)^2-(x^3)^2)[/itex] over the range between 0 and their intersection point.
 
  • #3
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1
12) Status is correct.

14) Integrate from 0 to 0.5 and multiply by 2Pi
15) Integrate both normally, multiply by 2Pi for the rotation, then subtract the integrals. Or equivalently, [tex] 2\pi \int{(x^3-25x)}{dx} [/tex]
 
  • #4
StatusX
Homework Helper
2,564
1
I'm sorry, I didn't see that you wanted the volume of the solid of rotation for 14. But I think for the solid of rotation you need to use pi r^2, not 2 pi, since you are adding up the areas of all the cross sections to get the volume.
 

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