# Find volume of a solid

dnt

## Homework Statement

tbe base of a solid is the region between y=x and y=x^2. find the volume of the solid if cross secions perpendicular to the x axis are isoceles right triangles with hypotenueses in xy plane.

n/a

## The Attempt at a Solution

honestly i cannot even mentally picture what this looks like, hence i cannot set up my integral. can someone help me understand what the question is even asking?

i know what the base looks like obviously but i cannot picture where these right triangles fit in.

## Answers and Replies

dnt
anyone? a little help?
thanks.

ChaoticLlama
I'll give you a hint.

Draw the graph of f(x)=x and g(x)=x² on the same grid. (i'm assuming here that the area is from 0 to 1)
Now define the height of this function as h(x) = f(x) - g(x).
h(x) is what you would integrate for area between two curves right?

Now off to the side, draw an isoceles triangle, and find the area in terms of the hypotanuse.

Then connect those two pictures together....

Because in this case, h(x) represents the hypotanuse of the triangle. And you need to imagine the graph you drew in the xy-plane has triangles extending up in the z-direction

good luck!

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dnt
hmm...am i understand this correctly:

so if i draw x and x^2 at each x point, the vertical line drawn from x to x^2 represents the hypoteneuse and the actual triangles come out of the page, and those two legs would be equal to each other? then you repeat that for all x values from 0 to 1?

edit: and therefore the actual right angle of each triangle would be in the air above the paper?

very difficult to mentally picture it...hope im saying it correctly.

dnt
so another thing id have to is integrate the area of a triangle (from 0 to 1) but the area function has to be in terms of the hypoteneuse (which would simply be the difference of the two graphs)...correct?

ChaoticLlama
Yup, you sound like you're on the right track.

Except, I'm not sure where you got "right triangle" from

Homework Helper
You are told that the cross sections are "isocelese right triangles with the hypotenuse in the xy-plane" Of course, if an isocelese right triangle has legs of length h, then it has hypotenuse $Y= \sqrt{2}h$ so, reversing that, $h= Y/\sqrt{2}$. Now, Y is the (vertical) distance between the two curves so it is easy to find Y and then h. Of course, the area of such a triangle is (1/2)h2. Multiply that by the "thickness" of each triangle, dx, and integrate.You are told that the cross sections are "isocelese right triangles with the hypotenuse in the xy-plane" Of course, if an isocelese right triangle has legs of length h, then it has hypotenuse $Y= \sqrt{2}h$ so, reversing that, $h= Y/\sqrt{2}$. Now, Y is the (vertical) distance between the two curves so it is easy to find Y and then h. Of course, the area of such a triangle is (1/2)h2. Multiply that by the "thickness" of each triangle, dx, and integrate.

dnt
Yup, you sound like you're on the right track.

Except, I'm not sure where you got "right triangle" from

the problem says "isoceles right triangles"

dnt
You are told that the cross sections are "isocelese right triangles with the hypotenuse in the xy-plane" Of course, if an isocelese right triangle has legs of length h, then it has hypotenuse $Y= \sqrt{2}h$ so, reversing that, $h= Y/\sqrt{2}$. Now, Y is the (vertical) distance between the two curves so it is easy to find Y and then h. Of course, the area of such a triangle is (1/2)h2. Multiply that by the "thickness" of each triangle, dx, and integrate.You are told that the cross sections are "isocelese right triangles with the hypotenuse in the xy-plane" Of course, if an isocelese right triangle has legs of length h, then it has hypotenuse $Y= \sqrt{2}h$ so, reversing that, $h= Y/\sqrt{2}$. Now, Y is the (vertical) distance between the two curves so it is easy to find Y and then h. Of course, the area of such a triangle is (1/2)h2. Multiply that by the "thickness" of each triangle, dx, and integrate.

I worked on this last night before reading your post: let me know if this is the same thing and if im on the right track.

my integral is from 0 to 1 of the function [(x-x^2)^2]/4

i got that because the area of a triangle is obviously 1/2 bh, but both my b and my h are (x/square root of 2) where x is the hypotenuse. but since my hypotenuse is just the difference of the two functions i substituted in x-x^2. how does that sound?

(i really need to learn how to make those functions look nice on this forum...)

pki15
You're correct, now multiply out and integrate! 