Disc method for volumes
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Disc method for volumes

[From: ] [author: ] [Date: 11-05-27] [Hit: ]
Can someone please show me step-by-step how to do each of these?Find the volume of the solid whose base is bounded by the graphs of y=x+1 and y=x^2-1, with the indicated cross sections taken perpendicular to the x-axis.(a) Rectangles of height 1.Im also given a graph and on this graph, there is a skinny rectangle placed on the x-axis where x=1.......
In Calculus, I missed a couple of days when we went over this in class and although my teacher explained it to me, I'm still a little confused. I've been going at these problems for a while, but I just can't seem to get them right. Can someone please show me step-by-step how to do each of these? >< I'd really appreciate it!


The first one says:

Find the volume of the solid whose base is bounded by the graphs of y=x+1 and y=x^2-1, with the indicated cross sections taken perpendicular to the x-axis.
(a) Rectangles of height 1.
I'm also given a graph and on this graph, there is a skinny rectangle placed on the x-axis where x=1.



And the last one:

Find the volume of the solid whose base is bounded by the circle x^2+y^2=4 with the indicated cross sections taken perpendicular to the x-axis.
(C) Semicircles
I'm also given a graph of a 3d semicircle that is basically centered on the origin with some endpoints at x=2 and y=2.




Thank you! You're really a lifesaver!

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1) Graph the equations y=x+1 and y=x^2-1; they intersect at x=-1 and x=2;

y = (x+1)-(x^2-1) = x-x^2+2

Area of rectangle = length x width = (x-x^2+2)(1) = x-x^2+2

Integrate between -1 and 2
∫ (x-x^2+2) dx from -1 to 2

= x^2 /2 - x^3/3 + 2x from -1 to 2
Let F(x) = x^2 /2 - x^3/3 + 2x
F(2)-F(-1) = 3.3333- (-1.1667)
=4.5

2)
y^2=4-x^2
y=sqrt(4-x^2)

radius = y
area = pi y^2 = pi(4-x^2)

Integrate pi(4-x^2) from -2 to 2

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hey. this website i've wrote on the source could perhaps be remarkable in your case
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keywords: method,for,volumes,Disc,Disc method for volumes
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