Volume of Solid Bounded by Circle and Plane

[tix24]

Homework Statement

find the volume of the solid based on the interior of the circle, r=cos(theta), and capped by the plane z=x.

Homework Equations

The Attempt at a Solution

i have drawn out the circle of equation r=cos(theta). I think that since z=x and is above the region, we have to use the double integral over this circular region. and integrate the function f(x,y)=x which in polar coordinate for would be rcos(theta).

so my train of thought is the following:

(∫ dtheta )(∫ (rcos(theta))rdr
where the limits of integration are for ∫ dtheta 0 to 2π
for ∫ (rcos(theta)rdr are from 0 to rcos(theta)

im not sure if my integrals are set up correctly any help regarding this problem would be very much appreciated. I have been stuck on this for a while

 

[vela]

It looks good except for your limits on the $\theta$ integral.

 

[tix24]

So how would I go about finding the limits for the r integral? I also think that the function which I'm integrating is incorrect.

 

[vela]

I didn't notice the mistake in the limits on the $r$ integral as well.

Suppose you wanted to calculate the area of the circle. What integral would you set up for that?

 

[tix24]

I didn't notice the mistake in the limits on the $r$ integral as well.

Suppose you wanted to calculate the area of the circle. What integral would you set up for that?

to calculate area by double integral i would have to integrate over the constant 1; my bounds of integration would be from r=0 to r=cos(theta) and i would have the second integral from theta=-Pi/2 to theta=Pi/2 for the theta integral

 

[vela]

Good. Now to get the volume, instead of integrating the function 1, you want to integrate the height of the solid as a function of $r$ and $\theta$.

 

[tix24]

So since the circle is capped be the plane, we have to integrate over the function f (x,y)=x which in polar coordinates is rcos (theta). I'm still not a 100% sure if this is correct or not.

 

[haruspex]

So since the circle is capped be the plane, we have to integrate over the function f (x,y)=x which in polar coordinates is rcos (theta). I'm still not a 100% sure if this is correct or not.

You're doing fine. What integral do you get?

 

[tix24]

So I set it up the following way:
The integral for theta went from -pi/2 to pi/2 and the integral for r went from 0 to cos (theta) the function which I integrated was rcos (theta) the result was pi/8

Can anybody confirm this?

 

[haruspex]

So I set it up the following way:
The integral for theta went from -pi/2 to pi/2 and the integral for r went from 0 to cos (theta) the function which I integrated was rcos (theta) the result was pi/8

Can anybody confirm this?

Doesn't sound right, though I've not checked in detail. I meant, what integral expression do you get?

 

[tix24]

my

Doesn't sound right, though I've not checked in detail. I meant, what integral expression do you get?

my integral expression was as follows: integral dtheta (bounds from -pi/2 to pi/2) integral rdr (bounds from r=0 and r=cos(theta) and the expression which i integrate is x. which i wrote as x=rcos(theta)

 

[haruspex]

mymy integral expression was as follows: integral dtheta (bounds from -pi/2 to pi/2) integral rdr (bounds from r=0 and r=cos(theta) and the expression which i integrate is x. which i wrote as x=rcos(theta)

Ok, that's correct, and I do get pi/8. I hadn't expected the 1/3 to get cancelled.