A hemispherical salad bowl has radius r we place an iron


1. A hemispherical salad bowl has radius R. We place an iron ball with radius r inside the bowl. Assume that 2r ≤R. Then we pour water into the bowl until the height of water is h.

(a) Assume 0 ≤ h 2r. Calculate the total volume of water.

(b) Assume 2r ≤ h R. Calculate the total volume of water.

Suggestions: You can check your solution! Your two answers should agree when

h = 2r. Also, you know what your answer should be when h = R.

2. Two long cylinders have radii r and R respectively (with r ≤ R) and their axes meet at a right angle. We want to compute the volume V of their intersection.

(a) Set up an integral whose value equals V. Leave your answer indicated as an integral.

Note: For those interested, there isn't any "nice" antiderivative for this function. Specifically, we can prove that the antiderivative we want is not an elementary function.

(b) Let a = R/r. Use a substitution to show that V = r3 F (a). In other words, we can write the volume V as r3 times an expression that depends only on a (and not on R or r separatedly). Write a formula for F.

Note: This result can be interpreted geometrically via scaling. Imagine that we multiply the radii of both cylinders by a factor of k (thus leaving a unchanged).

Convince yourself that the new volume should be exactly k3 times the old volume. That is what V = r3 F (a) means.

Hint: You can check your final answer: visit Wolfram Alpha online and ask it to compute the value of F (2). It should be approximately 12.16.

Note: This question can be solved in various ways and your grader is not a mindreader. Make sure to present your answer in a clean, organized, and well-explained manner. If your grader cannot understand what you are doing, you will get no credit, even if your answer is correct.

Solution Preview :

Prepared by a verified Expert
Mathematics: A hemispherical salad bowl has radius r we place an iron
Reference No:- TGS01349228

Now Priced at $30 (50% Discount)

Recommended (96%)

Rated (4.8/5)