Find the volume of the solid

[MarkSA]

Hi,

1) Use the double iterated integral to find the volume of the solid bounded by surface x + z^2 = 1, and y=x, and x=y^2

I'm having a lot of trouble figuring out limits on these problems. I don't really know how to graph the 3d surfaces.. my teacher never taught it and now it's coming back to haunt me. if anyone knows of a website that can teach it, calculator program for TI89, or anything to help i would be very grateful.

In the x-y plane I have y=x and x=y^2. This looks like a parabola opening to the right, with that and a straight line defining its bounds. And for my function I have z = sqrt(1-x)

\(\displaystyle \int^{1}_{0}\int^{y}_{y^2} \sqrt{1-x} \ dxdy\)

When I solve this I get pi/8 - 4/15. This is wrong. The correct answer is exactly 2x that value. I can't see where I could be going wrong in this though. Any ideas?

 

[galactus]

You have a square root, wouldn't you just multiply by 2?.

\(\displaystyle z=\pm\sqrt{1-x}\)

 

[Deleted member 4993]

You have a square root, wouldn't you just multiply by 2?.

\(\displaystyle z=\pm\sqrt{x-1}\)

If the function was:

\(\displaystyle z^2 \, = \, 1\, - \, x\)

Then Glactus is correct (sort of).

 

[MarkSA]

Why must I multiply it by 2 with the square root? In the x-y plane, the parabola and the line only seem to intersect at 2 points, seemingly making only one bounded region. Does it have something to do with x + z^2 = 1 being symmetrical about the x-y plane (is it?) Problem is I only have a vague idea of what that surface looks like.. just that it is probably some sort of paraboloid, opening along the y-axis?

 

[galactus]

Let's do it with a triple integral then. The radical is symmetric, as you stated. I fixed my typo.

\(\displaystyle \int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{-\sqrt{1-x}}^{\sqrt{1-x}}dzdydx=\frac{\pi}{4}-\frac{8}{15}\)

See what I mean?. See the upper and lower limits in the z integral of the above?. That shows what I mean.

In the double, just multiply by 2.

 

[Deleted member 4993]

Why must I multiply it by 2 with the square root? In the x-y plane, the parabola and the line only seem to intersect at 2 points, seemingly making only one bounded region. Does it have something to do with

x + z^2 = 1 being symmetrical about the x-y plane (is it?)

When you have even power (only) of one of the co-ordinate axis- the resulting graph (surface, volume) will be symmetric w.r.t. the plane (the axis) it is perpendicular.

Since you have z[sup:3evf2jt3]2[/sup:3evf2jt3] term (only) - the graph is symmetric to z axis.




Problem is I only have a vague idea of what that surface looks like.. just that it is probably some sort of paraboloid, opening along the y-axis?

 

[MarkSA]

thanks, I understand now where the 2x is coming from. i see that the integral i was using just found the volume of the portion above the z axis (as if it was bounded by z=0)