Calculation of volume of solid

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MathsLearner

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I am trying to solve the above problem. As per my thinking the diagram is some thing like this
1589099925059.png
the question says perpendicular to x - axis, there are two possibilities along the y-axis or along the z-axis, I think it is the z-axis. I am not sure how the solid looks like. Are there any on line tools to draw these diagrams?
My calculations are Area of the triangle is
[MATH] \frac 1 2 * 1 * 1 = 0.5 \\ V = \int A(x) dx [/MATH]I am not sure of the limits of the integral and the area in terms of x. Please help.
 
The cross-sections are formed by planes parallel to the yz plane! They are squares, not just lines.

You don't need to be able to draw the shape. First write the equation of the line; y will be the length of each side of the square cross-section at location x. That will give you the integrand.

What you should draw is an element of area, a rectangle with base delta x and height y, a function of x. Each such element will correspond to an element of volume, a square prism. If you do that, you should see what the limits of integration are; they are the same as for finding the area of the region.

The area of the triangle itself is irrelevant.
 
No, the side of each square is y, not x, so your integrand is not what it should be. What is y for a given value of x? That is, what is the equation of the line through (0,1) and (1,0)?

Now, your answer is correct, and your integral would be correct if you placed your coordinate system differently. But that, as far as I can tell, is an accident.
 
MathsLearner said:
Ok i think it is then
[MATH] \int_0^1 x^2 = [\frac {x^3} 3]_0^1 = \frac 1 3 [/MATH]
Click to expand...
The sizes of your squares are going the wrong direction. At x = 0, the square should have a side length of 1.

You may get the correct result, but only by accident.

Note: What are "a" and "b" doing in there? Very confusing.
 
Sorry for lacking the basics, this is the first time i am practicing these problems to solve some of the physics problems (Electro magnetism) which includes spherical related problems which need the knowledge of areas and volumes etc (though not very depth). Finally i think i could imagine the figure the solution i hope is
[MATH] \int_0^1 y^2 dx [/MATH]. The straight line equation is [MATH] y = -x + 1; [/MATH][MATH] \int_0^1(1-x)^2 dx[/MATH]The answer is [MATH] \frac 1 3[/MATH]. Am i correct now?
 
Before you do these type problems you really should draw at least one 'slab' and find the volume of it. Then add them all up!
 
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