A is OK
B is
v = ∫[0,4] 2πrh dx, where
r = 4-x
h = (16-2x)-√x^3
For C, you need to split the Region into two parts at y=8, because we have to integrate along dy. The rectangles have a base of x, which changes at y=8
v = ∫[0,8] ∛y^2 h(y) dy + ∫[8,16] (8-1/2y) dy
The functions f and g are given by f(x)=sqrt(x^3) and g(x)=16-2x. Let R be the region bounded by the x-axis and the graphs of f and g.
A. Find the area of R.
B. The region R from x=0 to x=4 is rotated about the line x=4. Write, but do not evaluate, an integral expression the volume of the solid.
C. The region R is the base of a solid. For each y, the cross section of the solid taken perpendicular to the y-axis is a rectqngle whose base lies in R and whose height is h(y). Write, but do not evaluate, an integral expression of the volume of the solid.
A. I took the integral of f(x) with bounds 0 to 4 plus the integral of g(x) with bounds of 4 to 8. My area was 144/5 u^2.
B. The integral from 0 to 4 of pi(sqrt(x^3))^2 dx and then simplified the square and square root.
C. I solved each of the functions for x. Then set the integral up as the integral from 0 to 8 of [(8-1/2y)-(cubed root(y^2))] h(y) d(y)
Thank you for checking
1 answer