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A bowl is shaped like a hemisphere with diameter 30 $\mathrm{cm} . \mathrm{A}$

ball with diameter 10 $\mathrm{cm}$ is placed in the bowl and water is

poured into the bowl to a depth of $h$ centimeters. Find the

volume of water in the bowl.

$$

\begin{array}{l}{V(y)=\pi \cdot\left(15 h^{2}-\frac{1}{3} h^{3}-\frac{2500}{3}\right) \text { for } h \geq 10, \text { and } V(y)=10 \pi \cdot h^{2} \text { when }} \\ {h<10}\end{array}

$$

Applications of Integration

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Idaho State University

{'transcript': "So we have different formulas to the volume. It's gonna change. So the first part, when the ball is not fully submerged and then we have a second part when the ball is fully submerged. So we're going to use the washer method for H smaller than 10, and we have to express the bowl and the ball as a function of X as we revolve around the Y axis. So from the washer method, the outside diameter is going to be X squared, plus why minus 15 squared equals 15 square. So solving for X we get the X is equal to the square root of 225 minus y minus 15 squared and then the inner diameter is going to be given by X squared plus y minus five squared equals five squared. So when we saw for X, we end up getting the square root of 25 minus y minus five squared. Um, then the formula for the area is given as a of y, which is going to be equal to, um 20. Hi. Why so now what we have to do is just integrate from zero to H. So I have V of H being equal to pi times the integral from zero to h of 20 y Do you y that's going to give us 10 y squared, evaluated as your own h. So our final answer is going to be 10. Hi, h squared. Um, Now an H becomes larger than 10. The ball is fully submerged. So the volume added from that point can be calculated by plugging in H equals 10. Sophie, plug in H equals 10. What we end up getting is 1000 pie. Um, the area of disks above H equals 10 can be calculated using the next equation. So in that case, what we'll have is that a of Y is equal to pi times 30 y minus y squared. And that's compared to this simpler equation that we had before. So now we're just calculate this volume as a function of why it's gonna be pi times the integral from 10 to h of 30 y this time minus y squared. Do you? Why, once this is simplified, we end up getting this volume is pi Times 15 h squared minus one third times h cubed minus 2500 over three"}

California Baptist University

Applications of Integration