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Entries from March 2008

Chapter 8.1, Problem 35

March 25, 2008 · Leave a Comment

Find the surface area obtained by rotating the curve y=(4-x^{2/3})^{3/2} around the x-axis over the interval [0,8]

Answer: y'=-x^{-1/3}(4-x^{2/3})^{1/2}, so (y')^2 = \frac{4-x^{2/3}}{x^{2/3}}.

1+(y')^2 = 4x^{-2/3}

y \sqrt{1+(y')^2} = (4-x^{2/3})^{3/2}2x^{-1/3}

The surface area is therefore 2\pi \int_0^8 (4-x^{2/3})^{3/2}2x^{-1/3}dx. Make the u-substitution u = 4-x^{2/3} and the integral becomes 6\pi\int_0^4 u^{3/2} du, which can be easily computed, and which I get works out to \frac{384\pi}{5}.

Categories: Math 132 Spring 2008

Test

March 11, 2008 · Leave a Comment

This is plain text. Next up is math.

\int\left|fg\right|d\mu \leq (\int |f|^p d\mu)^{\frac{1}{p}}(\int |g|^q d\mu)^{\frac{1}{q}}

Testing one, two three.

Categories: Pay no attention to the man behind the curtain