Volume of cross sections using rectangles

  • #1

Homework Statement



The area bounded by y=√x, the x axis and the line x=9 and is perpendicular to the x axis. find the volume of the cross section using rectangle with the h=1/2b

Homework Equations





The Attempt at a Solution


I just want to know if i'm correct or on the right track. if not please correct me

A=HB
A=(1/2B)B =3/2B
So i take the integral of 3/2B from 0 to 9
V=3/2 integral 0 to 9 B
V=3B^2/4 from 0 to 9
V=243/4
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



The area bounded by y=√x, the x axis and the line x=9 and is perpendicular to the x axis. find the volume of the cross section using rectangle with the h=1/2b

Homework Equations





The Attempt at a Solution


I just want to know if i'm correct or on the right track. if not please correct me

A=HB
A=(1/2B)B =3/2B
So i take the integral of 3/2B from 0 to 9
V=3/2 integral 0 to 9 B
V=3B^2/4 from 0 to 9
V=243/4

You haven't stated the problem clearly enough to be sure what it actually is, but I'm pretty certain your solution is wrong anyway.

Does the solid you are trying to describe sit on the described area with rectangular cross-sections perpendicular to the xy plane? Are the bases of the rectangular cross-sections in the xy plane perpendicular to the x-axis?
 
  • #3
You haven't stated the problem clearly enough to be sure what it actually is, but I'm pretty certain your solution is wrong anyway.

Does the solid you are trying to describe sit on the described area with rectangular cross-sections perpendicular to the xy plane? Are the bases of the rectangular cross-sections in the xy plane perpendicular to the x-axis?

Yes, it does sit on the area with rectangular cross sections perpendicular to the x and y plane and the bases of the rectangular cross section are perpendicular to the x axis. I just need to find the volume using the given information.

Also i realize an error in my work :
A=HB
A=(1/2B)B =3/2√ x
So i take the integral of 3/2√x from 0 to 9
V=3/2 integral 0 to 9 x^3/2/(3/2)
V=6x^3/2/(6) from 0 to 9
V=162/6

or

The base is going to be sqrt(x). Therefore the height is 1/2sqrt(x), and the area is 1/2 x. The volume will be the integral from 0 to 9 of 1/2 x, which is 81/4.

i get two different answers
 
Last edited:
  • #4
LCKurtz
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Yes, it does sit on the area with rectangular cross sections perpendicular to the x and y plane and the bases of the rectangular cross section are perpendicular to the x axis. I just need to find the volume using the given information.

Also i realize an error in my work :
A=HB
A=(1/2B)B =3/2√ x
So i take the integral of 3/2√x from 0 to 9
V=3/2 integral 0 to 9 x^3/2/(3/2)
V=6x^3/2/(6) from 0 to 9
V=162/6

or

The base is going to be sqrt(x). Therefore the height is 1/2sqrt(x), and the area is 1/2 x. The volume will be the integral from 0 to 9 of 1/2 x, which is 81/4.

i get two different answers

That's because (1/2)B*B doesn't equal (3/2)B.
 

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