How do you find the volume of the solid generated by revolving the region enclosed by the parabola #y^2=4x# and the line y=x revolved about the y-axis?

Question

Charles Arocha · Accepted Answer

#(128pi)/15#

Consider an horizontal strip of thickness #delta y# revolved around y axis. The volume of the solid so generated would be #pi x_1^2 delta y - pi x_2^2 deltay#. This is illustrated in the figure below.

The volume of the solid generated by revolving the who region bounded by the given parabola and the straight line would thus be

# int_(y=0) ^4 pi x_1^2 delta y - pi x_2^2 deltay#. Substituting #y= x_1# and #y^2 /4 = x_2#, te given integral would become

# pi int_0^4 (y^2 -y^4 /16 ) dy #

#pi [ y^3 /3 -y^5 /80]_0^4#

#pi[64/3 -1024/80]# = #64pi(1/3 -1/5)#= #(128pi)/15#

William Abdalla · Answer

undefined To find the volume of the solid generated by revolving the region enclosed by the parabola $y^2=4x$ and the line $y=x$ revolved about the y-axis, you can use the method of cylindrical shells. The volume can be calculated using the formula:

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

where $f(x)$ represents the height of the shell at a given x-value, and $a$ and $b$ are the x-values at the points of intersection of the parabola and the line.

First, find the intersection points of the parabola and the line by substituting $y=x$ into the equation of the parabola $y^2=4x$. This gives $x^2=4x$, which simplifies to $x(x-4)=0$, so $x=0$ and $x=4$.

Now, the height of the shell is the difference between the y-values of the parabola and the line, which is $f(x) = y_{\text{parabola}} - y_{\text{line}}$. Substituting the equations of the parabola and the line, we get $f(x) = \sqrt{4x} - x$.

Finally, integrate $x \cdot f(x)$ from $x=0$ to $x=4$ and multiply the result by $2\pi$ to find the volume of the solid. This integration will yield the volume of the solid generated by revolving the region about the y-axis.