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Numerade Educator

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Problem 49 Hard Difficulty

Find the volume of the described solid $ S $.
A cap of a sphere with radius $ r $ and height $ h $.

Answer

$\frac{1}{3} \pi h^{2}(3 r-h)$

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Video Transcript

okay. We know the equation of the circle with the center J K and radio. Are is X minus J squared plus y minus case where which is equivalent to our square. We also know the formula for volume is easy. As pie comes the integral from zero h Y squared DX and recalls the problem mistrust fighters were rotating to a right of the Y axis at the center of H minus or commas here. Therefore, given this, we end up with maximize H plus our which, as I specified earlier, is going to be in parentheses. And remember how we said the why is squared. So we see the red eye of the disks from rotating. It squared like I said, which means then we can expand this and we can integrate, which means we can use the power rule. Which means he increased the exponents by one divide by the new expert, so may increase the export to three. When we integrate, divided by the new expert of three, you could see this is what we did over here. We had an excellent the r squared. We end up with this again, we can pull out the Constance. 1/3 is a constant. Therefore it's been pulled out over here. Therefore, we now end up with three R squared times H minus. R. Cubed plus princes are minus h cute. No, an algebra problem. The foil method, which is essentially where you are able to expand. So, for example, or minus h can be all right. Each cube can be expanded using the oil. Nothing. And the distribution method means that we can multiply 1/3 pied times each of these individual terms. There's two of them, plus the foiled term over here. Okay, now that we have this, we know we can pull out 1/3 h squared because again, we know H squared is common within these terms. Therefore, could be pulled out on the outside. The pie doesn't go anywhere. It's been there since the beginning, and then we have three are minus h