Volume of Solid w/ Semicircular Cross Sections in 1st Quadrant

AI Thread Summary
The discussion revolves around calculating the volume of a solid with semicircular cross sections in the first quadrant, bounded by the x-axis, y-axis, and the line x + 2y = 8. Participants initially miscalculated the volume by incorrectly assuming the diameter of the semicircles. The correct approach involves determining the radius as (4 - x/2)/2, leading to the proper integral setup. The final volume is obtained by integrating from 0 to 8 using the formula 1/2π[(4 - x/2)/2]^2. Clarifications helped resolve the misunderstanding, confirming the correct answer aligns with choice c from the provided options.
radtad
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The base od a solid is a region in the 1st quadrant bounded by the x-axis, y-axis and the line x+2y=8. If cross sections of the solidperpendicular to the x-axis are semicircles, what is the volume of the solid?

How come the answer isn't just the intgegral from 0-8 of 1/2pi(4-x/2)^2
 
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radtad said:
The base od a solid is a region in the 1st quadrant bounded by the x-axis, y-axis and the line x+2y=8. If cross sections of the solidperpendicular to the x-axis are semicircles, what is the volume of the solid?

How come the answer isn't just the intgegral from 0-8 of 1/2pi(4-x/2)^2

That's what it looks like to me. Is it the wrong answer?
 
yea according to the answer key its wrong
 
choices are
a. 12.566 b. 14.661 c. 16.755 d 67.021 e 134.041

i keep ending up with choice d but the answer key says its choice c
 
Oh I see now. I misunderstood the problem. I did it by finding the solid after rotating about the x-axis and integrating... But the cross section of this solid is the circle whose diameter is from y=0 to y=4-x/2. We were thinking the cross section was a circle with diameter from y=-(4-x/2) to y=4-x/2. But if we make the radius (4-x/2)/2, then we get the right answer.

0-8 of 1/2pi[1/2(4-x/2)]^2 is correct.
 
yea i realized that too thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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