Volume of a bowl (using variables not numbers)

htoor9
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Homework Statement



A hemispherical bowl of radius "a" contains water to a depth "h"

Fine the volume of water in the bowl.


Homework Equations



Pi * Integral(R(x))^2

basically the disk method


The Attempt at a Solution



The answer is (Pi*h2*(3a-h))/3...i just have no idea how to set up the integral to get that. Thanks guys
 
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htoor9 said:

Homework Statement



A hemispherical bowl of radius "a" contains water to a depth "h"

Fine the volume of water in the bowl.


Homework Equations



Pi * Integral(R(x))^2

basically the disk method


The Attempt at a Solution



The answer is (Pi*h2*(3a-h))/3...i just have no idea how to set up the integral to get that. Thanks guys
Start by sketching a graph of the function that represents the bowl. I recommend sketching the lower half of the graph of x2 + y2 = a2 to make your computations a little simpler.

Draw a line across the bowl representing a level of h units. Since h is measured from the bottom of the bowl, if you have located the bowl as I suggested, the water level crosses the y-axis at (0, h - a). For example, if the bowl's radius is 6'' and the water is 2" deep, the water level on the y-axis is at (0, 2 - 6) = (0, -4).

Draw a typical "slice" of the water. What is its volume? That's what you will use for your integrand.
 
Mark44 said:
Start by sketching a graph of the function that represents the bowl. I recommend sketching the lower half of the graph of x2 + y2 = a2 to make your computations a little simpler.

Draw a line across the bowl representing a level of h units. Since h is measured from the bottom of the bowl, if you have located the bowl as I suggested, the water level crosses the y-axis at (0, h - a). For example, if the bowl's radius is 6'' and the water is 2" deep, the water level on the y-axis is at (0, 2 - 6) = (0, -4).

Draw a typical "slice" of the water. What is its volume? That's what you will use for your integrand.

since I'm revolving it around the y axis, how do i know what the length of the axis to the end of the bowl with the water is? I'm really confused
 
The bottom point of the bowl is at (0, -a).
 
Mark44 said:
The bottom point of the bowl is at (0, -a).

ok. i still can't figure out what the equation of the curve is... is it h^2?
 
An equation has = in it. Your graph should show the lower half of a circle whose center is at the origin, and whose radius is a. You know the equation of a circle, right?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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