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## Cross-sectional volume question?

Limits, differentiation, related rates, integration, trig integrals, etc.
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### Cross-sectional volume question?

by Thatguy73 on Wed Aug 18, 2010 2:07 am

So I would like to know if I am doing this correctly. Here's the question:

Solid B has a volume of 8(pi)

It also is a Pyramid with a square base and a height of 8. What is the length of the side (s) of the base of B?

The Cross-sectional area would be A(x)=s^2 And I would need to write s in terms of x to put it in the definite integral to find the volume.

Since the height of the pyramid 8, I can make the Vertex of it at (0,0) and then make the base centered at (8,0). So the cross-sectional areas would need to be summed up on the interval [0,8]. I can find the volume of the pyramid:

$\displaystyle V=8\pi=\int_{\0}^8 A(x)dx$

$A(x)=s^2$

At the end of the base where x=8, the side (s) of the base would be 2y. So then something times x must give 2y

$kx=2y=s$ Therefore $A(x)=(kx)^2=k^2x^2$

$\displaystyle V=8\pi=\int_{\0}^8 k^2x^2dx$ Since k^2 is a constant:

$8\pi=k^2\int_{\0}^8x^2dx$

$8\pi=k^2(8^3/3)$
Solving for k:

$k=0.384$

Since: $kx=s$ Plug in x=8 and k.

$s=3.070$

Could someone tell if this method is correct? I just started learning this and it's kind of confusing.

Thanks
Thatguy73

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### Re: Cross-sectional volume question?

by Martingale on Wed Aug 18, 2010 3:51 am

Thatguy73 wrote:So I would like to know if I am doing this correctly. Here's the question:

Solid B has a volume of 8(pi)

It also is a Pyramid with a square base and a height of 8. What is the length of the side (s) of the base of B?

The Cross-sectional area would be A(x)=s^2 And I would need to write s in terms of x to put it in the definite integral to find the volume.

Since the height of the pyramid 8, I can make the Vertex of it at (0,0) and then make the base centered at (8,0). So the cross-sectional areas would need to be summed up on the interval [0,8]. I can find the volume of the pyramid:

$\displaystyle V=8\pi=\int_{\0}^8 A(x)dx$

$A(x)=s^2$

At the end of the base where x=8, the side (s) of the base would be 2y. So then something times x must give 2y

$kx=2y=s$ Therefore $A(x)=(kx)^2=k^2x^2$

$\displaystyle V=8\pi=\int_{\0}^8 k^2x^2dx$ Since k^2 is a constant:

$8\pi=k^2\int_{\0}^8x^2dx$

$8\pi=k^2(8^3/3)$
Solving for k:

$k=0.384$

Since: $kx=s$ Plug in x=8 and k.

$s=3.070$

Could someone tell if this method is correct? I just started learning this and it's kind of confusing.

Thanks

your answer is correct (if you allow for rounding)
the exact answer is $\sqrt{3\pi}$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

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