volume question + Simpson's rule

**Dr Zoidburg** · July 30th 2008, 10:38 PM

Two Questions, one I'm pretty confident I've done and just want confirmation, the other I'm stuck on:

First Question:
Use Simpson's Rule, n=4, to approximate:
$\int^5_2 \sqrt(x-1).dx$
and find a bound on the error.
I won't bore you all with the full result:
h=3/4
k=0, x0=2, y0=2, w0=1, yw=2
k=1, x1=2.75, y1=2.658, w1=4, yw=10.583
k=2, x2=3.5, y2=3.162, w2=2, yw=6.325
k=3, x3=4.25, y3=3.605, w3=4, yw=14.422
k=4, x4=5, y4=4, w5=1, yw=4
Simpson's approximation = 3/12*37.330 = 9.332 (3dp)
Hope that's correct!
Error:
$f''''(x)=\frac{-15}{16}(x-1)^{\frac{-7}{2}}$
x=5, this is -0.007.
Error = (-0.007*243)/(180*256) = -0.0000386
also correct?

Okay onto the volume question I'm stuck on:
Base of a solid V is the region bounded by $y=x^2$ and $y=2-x^2$ . Find the volume if V has square cross sections.
tbh, I'm really stuck on this one and haven't the foggiest!

**Chris L T521** · July 30th 2008, 10:46 PM

Originally Posted by Dr Zoidburg

Okay onto the volume question I'm stuck on:
Base of a solid V is the region bounded by $y=x^2$ and $y=2-x^2$ . Find the volume if V has square cross sections.
tbh, I'm really stuck on this one and haven't the foggiest!

Find the intersection points. They will define the upper and lower bounds.

Now for the volume.

Recall that $V(x)=\int_a^b A(x)\,dx$

Note the region that we are given.

We see that the width of the base is the distance between the 2 graphs:

$D=2-x^2-x^2=2-2x^2$

Since $A_{\square}=D^2$ , we see that $A(x)=(2-2x^2)^2$

Therefore, $V(x)=\int_{-1}^{1}(2-2x^2)^2\,dx$

Does this make sense?

--Chris

**Dr Zoidburg** · July 31st 2008, 08:58 PM

yes it does!
Thanks very much for that.
I'd got as far as finding the intersect points but no further. I missed your next step so when I tried integrating I got zero! D'oh!
Thanks again. That's another assignment finished, tg.

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