How do you find the volume of the solid generated by revolving the region bounded by the graphs of the equations #y=sqrtx#, y=0, and x=4 about the y-axis?

Question

Isabella Ackerman · Accepted Answer

V=#8pi# volume units

Essentially the problem you have is: V=#piint_0^4 ((sqrtx))^2 dx# Remember, the volume of a solid is given by: V=#piint (f(x))^2 dx# Thus, our original Intergral corresponds: V=#piint_0^4(x) dx# Which is in turn equal to: V=#pi [ x^2/(2)]# between x=0 as our lower limit and x=4 as our upper limit. Using The fundamental theorem of Calculus we substitute our limits into our integrated expression as subtract the lower limit from the upper limit. V=#pi[16/2-0]# V=#8pi# volume units

Christopher Agresta · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the graphs of the equations $ y = \sqrt{x} $, $ y = 0 $, and $ x = 4 $ about the y-axis, we use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

$$ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx $$

where $ f(x) $ is the height of the shell, and $ a $ and $ b $ are the bounds of integration.

In this case, since we're revolving about the y-axis, we need to express $ x $ in terms of $ y $ to set up the integral.

From $ y = \sqrt{x} $, we have $ x = y^2 $.

The bounds of integration are from $ y = 0 $ to $ y = 2 $ (since $ y = \sqrt{x} $ intersects $ x = 4 $ at $ y = 2 $).

Now, substituting into the formula, we get:

$$ V = 2\pi \int_{0}^{2} y^2 \cdot y \, dy $$

$$ V = 2\pi \int_{0}^{2} y^3 \, dy $$

$$ V = 2\pi \left[ \frac{1}{4}y^4 \right]_{0}^{2} $$

$$ V = 2\pi \left( \frac{1}{4} \cdot 2^4 - \frac{1}{4} \cdot 0^4 \right) $$

$$ V = 2\pi \left( \frac{16}{4} \right) $$

$$ V = 8\pi $$

So, the volume of the solid generated by revolving the region about the y-axis is $ 8\pi $ cubic units.