Help on finding volume of integral analytically

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thatguy47

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I was able to solve this on the calculator (I think its right) but I can't figure out how to solve it by hand. Here's the problem:
A solid lies between two planes perpendicular to the x-axis at x=1 and x=3. The cross-sections perpendicular to the x-axis are squares that have the sides running from the curve y=x^(3/2) down to the x-axis. Find an integral for the volume of the solid and solve analytically.

I first graphed it and got V = [integral 1 to 3] (x^(3/2))^2 dx
I plugged that into the calculator and got 1093/14 which I assume is the right answer. I just can't seem to get this by solving it analytically. Here's my steps for trying to solve it:
V = [integral 1 to 3] x^(9/4) dx >>>I squared x^(3/2) to get x^(9/4)
Next I did (4/13)x^(13/4) and solved that by subtracting x=1 from x=3 and I got a different answer than 1093/14.

What am I doing wrong?
 
I get 20 as the answer.

Here is your problem: \(\displaystyle (x^{\frac{3}{2}})^{2}=x^{3}\)

The exponent of an exponent is multiplied. You would multiply 3/2 and 2, not square 3/2.

\(\displaystyle \int_{1}^{3}x^{3}dx\)
 
Thanks, I can't believe I was squaring the exponents. I also entered the equation wrong in the calculator :roll: . I understand it now. Thanks for the help.
 
What if they were semicircles instead of squares?. Can you do that one?.
 
you've seen this before, remember?

diameter of the circle is \(\displaystyle d = x^{\frac{3}{2}}\)
\(\displaystyle r = \frac{d}{2}\)
\(\displaystyle r^2 = \frac{x^3}{4}\)

area of a cross section is \(\displaystyle A = \frac{\pi r^2}{2} = \frac{\pi}{2} \cdot\frac{x^3}{4}\)
 
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