Thread: computing area and volume with integrals

1.what is the volume of solid whose cross sections are squares perpendicular to the x- axis where the solid is based on the region bounded by the curve y= x^3 and the x axis and x=1?
<a>1/3
<b> 1/6
<c> 1/7
<d>1/4

2.What is the area of the region bounded by the curves y = x^4 + x^3 – 2x + 3 and y = x^3 + 4x^2 – 2x + 3?
<a> 0
<b> 64/15
<c> 128/15
<d> 372/15

Originally Posted by harry

1.what is the volume of solid whose cross sections are squares perpendicular to the x- axis where the solid is based on the region bounded by the curve y= x^3 and the x axis and x=1?
<a>1/3
<b> 1/6
<c> 1/7
<d>1/4

2.What is the area of the region bounded by the curves y = x^4 + x^3 – 2x + 3 and y = x^3 + 4x^2 – 2x + 3?
<a> 0
<b> 64/15
<c> 128/15
<d> 372/15

this is number 2, see the diagram below. we want the area of the grey shaded region.
What is the area of the region bounded by the curves y = x^4 + x^3 – 2x + 3 and y = x^3 + 4x^2 – 2x + 3


first we want to draw a diagram (below) then we want to find our limits of integration. we do that by finding where the curves intersect.

so y = x^4 + x^3 - 2x + 3 i'll call this f(x) and y = x^3 + 4x^2 - 2x + 3 i'll call this g(x)
=> x^4 + x^3 - 2x + 3 = x^3 + 4x^2 - 2x + 3
=> x^4 - 4x^2 = 0
=> x^2(x^2 - 4) = 0
=> x^2(x + 2)(x - 2) = 0
=> x = 0, x = -2 or x = 2 ...these are our points of intersection.
now we find whish graph is higher in both the regions [-2,0] and [0,2]. in both these regions, g(x) is greater than or equal to f(x), again, see diagram, if you could not produce the diagram, just plug in numbers in these regions into both f(x) and g(x), which ever one yields the greatest value is the higher function.

note that if differente functions were higher in the different intervals you would have to split the integral. example, if g(x) > f(x) on interval [-2,0] and f(x) > g(x) on [0,2], then the area would be
A = int{g(x) - f(x)}dx on [-2,0] + int{f(x) - g(x)}dx on [0,2]

since g(x) is always greater, we can integrate over the entire interval with one function, namely, g(x) - f(x)

so A = int{g(x) - f(x)}dx
=> A = int{x^3 + 4x^2 - 2x + 3 - (x^4 + x^3 - 2x + 3)}dx
=> A = int{-x^4 + 4x^2}dx
=> A = [(-1/5)x^5 + (4/3)x^3] evaluated between 2 and -2
=> A = -32/5 + 32/3 - 32/5 + 32/3
=> A = 128/15

Jhevon what program are you using to make these cool pictures?

Originally Posted by harry

1.what is the volume of solid whose cross sections are squares perpendicular to the x- axis where the solid is based on the region bounded by the curve y= x^3 and the x axis and x=1?
<a>1/3
<b> 1/6
<c> 1/7
<d>1/4

See diagram below. the cross-section are 1 x 1 squares. so we will find the volume by using the formula:

V = l*w*h = (Area)*h ........the height is 1 from above, so now all that remains is to find the area of the shaded region, then we times that by one to find the volume.

A = int{x^3}dx ....evaluated between 0 and 1
=> A = [(1/4)x^4] evaluated between 0 and 1
=> A = (1/4)(1)^4 - 0
=> A = 1/4

so V = (1/4)* 1 = 1/4

Originally Posted by qbkr21

Jhevon what program are you using to make these cool pictures?

I use this lovely program called "Graph", i use it to draw the graphs, then save the pic and copy it in paint, where i add shade and text. you can get it here Graph.

Oh...yea this program is SUPER SWEET!

Originally Posted by qbkr21

Oh...yea this program is SUPER SWEET!

I am the one who introduced this program to MHF


Funny, Matlab, a super un-user friendly program, extremely expensive, is not as great as this free one!

Ohh... I see you cannot simply click the save icon with this program. You must select File => Save as Image => Then select .gif or .jpg. This solves the problem

Originally Posted by qbkr21

Ohh... I see you cannot simply click the save icon with this program. You must select File => Save as Image => Then select .gif or .jpg. This solves the problem

correct. to save as an image that's the way to do it. just clicking save saves it as a graph file