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vol help! (1 Viewer)
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nightweaver066
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I think the answer is 16pi^2 (using theorem of Pappus)
Best to type up what you've done or take a photo so we can see if you've made a mistake or not.
hey i typed wrong radius value...sorry my bad...
nightweaver066
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Yep you're right then.
i dont think so...in the end i got 4pie * [(integral of)(lower limit 2, upper limit 6)] [(2+y)(sqroot (4-(y-4)^2) dy...am i goin correct??? what will b the next step?
also which method do u think is the best when calculating volume of torus, cylindrical or washers???
nightweaver066
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actually sorry guys, get back to you later, or maybe someone else can help out, busy atm
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asianese
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Indeed the answer is 
both by Pappus theorem and washers.
I had an impossible time using shells since there was an ugly (y-4)^2 so I tried washers, that run strips top to bottom [] like that along the circle.
^2=4-x^2\\y=4\pm\sqrt{4-x^2}\\ $Let the larger radius be R and the smaller radius be r. $ \\ dV=(R^2-r^2)\pi dx \\ = ((4+\sqrt{4-x^2})^2-(4-\sqrt{4-x^2})^2)\pi dx \\ = 16\pi \sqrt{4-x^2} dx \\ V=\lim_{dx \to 0} \sum_{x=-2}^2 16\pi \sqrt{4-x^2} dx \\ = 32\pi \int_0^2 \sqrt{4-x^2} dx $ Since $ \sqrt{4-x^2} $ is even. $\\ $ Set $ x=2\sin u \therefore dx=2\cos u du \\ V=32\pi \int_0^{\frac{\pi}{2}}4\cos^2u du \\ = 128\pi \int_0^{\frac{\pi}{2}} \cos^2u du)
Which gives
after evaluation.
Theorem of Pappus is something along the lines of, the volume swept out is equal to the area of a slice, multiplied by the distance the centre sweeps out.
I had an impossible time using shells since there was an ugly (y-4)^2 so I tried washers, that run strips top to bottom [] like that along the circle.
Which gives
Theorem of Pappus is something along the lines of, the volume swept out is equal to the area of a slice, multiplied by the distance the centre sweeps out.
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nightweaver066
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good solution asianese!
i should have posted sooner, but cylindrical shells works but the integration is uglier.
Cylindrical shells method:
 - y^2) \times 2x)
^2 - y^2))
^2 \approx 0)
^2} \delta y)
^2} \delta y)
^2}dy)
 \times 2cosu \times 2cosudu)
cos^2udu)
)
i should have posted sooner, but cylindrical shells works but the integration is uglier.
Cylindrical shells method:
Last edited:
asianese
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Yeah...I was reluctant to use washers (since I prefer shells) but it worked out better for this problem.
Note ppl that theorem of pappus isn't in the syllabus and you'll get no marks if you only say pir^2*2piR. It is, however, a very good way to check your answer!
Note ppl that theorem of pappus isn't in the syllabus and you'll get no marks if you only say pir^2*2piR. It is, however, a very good way to check your answer!