finding the perimeter and volume using calc

**michelle91** · May 14th 2009, 08:06 PM

ok so i really do not understand what i have to do here and i would love if someone could help me out just a bit:
i have to find the perimeter and volume of a solid.the base od the solid is the region of the xy-lane bounded by: y=x^2 -2x +2 , y=0, x=0, and x=3.
the cross section is a semicircle with diameter in the base of the solid.
does anyone have any idea as to how i should go about doing this?thanks in advance =)

**redsoxfan325** · May 14th 2009, 09:50 PM

Originally Posted by michelle91

ok so i really do not understand what i have to do here and i would love if someone could help me out just a bit:
i have to find the perimeter and volume of a solid.the base od the solid is the region of the xy-lane bounded by: y=x^2 -2x +2 , y=0, x=0, and x=3.
the cross section is a semicircle with diameter in the base of the solid.
does anyone have any idea as to how i should go about doing this?thanks in advance =)

For the volume, you want first to find what the radius of your semicircles are, because you want to integrate $\frac{\pi r^2}{2}\,dx$ . We know that $r=\frac{1}{2}f(x)$ (because $f(x)$ is the diameter, so we need to divide by two to get the radius), so we want to integrate $\frac{\pi\cdot\frac{1}{4}f^2(x)}{2}$ over $[0,3]$ .

$\frac{\pi}{8}\int_0^3 (x^2-2x+2)^2\,dx$

Multiply it out and integrate. Final answer:

Spoiler:

And what do you mean by perimeter of a solid. Do you mean surface area or perimeter around the base?

**michelle91** · May 15th 2009, 03:19 AM

yes for the perimeter i meant to say surface area, sorry..btu as for the volume thank u so very much!!!

**Pinkk** · May 15th 2009, 05:34 AM

You would have to integrate the arc length of the semi-circle cross section over $[0,3]$ . The arc length of a semi-circle is half the circumference of a circle, or $\frac{\pi d}{2}$ , and in this case, $d=f(x)$ . So the intergral would be:

$\frac{\pi}{2}\int_{0}^{3}x^{2}-2x+2\,\,dx$

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