Question

R is in the first quandrant enclosed by y=(square root of) 6x+4, the line y=2x, and y-axis.
1. What is the are of R?
2. What is the integral expression of a single variable of the volume of the solid when R is revolved about the x-axis?
3. What is the integral of a single variable for the volume of the solid when R is revolved about the y-axis?
*please answer each part and how you got it

Answer 1

***what is the area

Answer 2

please help

Answer 3

To find area of R, integrate from 0 to 2 and take area below sqrt function minus area below line y=2x
\[A=\int\limits_{0}^{2}(\sqrt{6x+4} -2x) dx\]
To find volume by revolving around x-axis, use disc method
each vertical cross-section is a circular ring with inner radius (r) and outer radius (R)
area of ring = pi(R^2 -r^2)
R is function furthest from x_axis, sqrt function
r is y=2x
\[V = \pi \int\limits_{0}^{2}(6x+4) -4x^{2}\]
To find volume by revolving R around y_axis, use shell method
sum up all the surface areas from center to outer perimeter
surface area = 2*pi*radius*height
radius =x, goes from 0 to 2
height is the difference of 2 functions, sqrt(6x+4) -2x
\[V = 2 \pi \int\limits_{0}^{2}x(\sqrt{6x+4} -2x) dx\]