(y2-y1) dx
but y1 is the x axis so y1 = 0
so I get integral from - 1 to + 1 of (x+1) dx = x^2/2 + x at 1 - at -1
The base of a solid is bounded by the curve y=sqrt(x+1) , the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid
a. 1
b. 2
c. 2.333
d. none of the above
I got a little confused, but this is what I have so far. Can someone help me get my work straight?
V=lwh so V=side^2 dy = the integral from -1 to 1 of (y2-y1) dx = (sqrt(x+1)-1)^2 dx = the integral from -1 to 1 of x+1+1 dx = x^2/2 + 2x [-1,1] = (1^2/2+2(1))-(-1^2/2+2(-1)) = 1/2+2-(1/2-2) = 5/2-(-3/2) = 4?
3 answers
I mean'(y2-y1)^2 dx = (x+1) ddx
The answer should be 2.
3 answers