# Thread: Is this problem solved correctly? (integrals)

1. ## Is this problem solved correctly? (integrals)

Question:
Find a formula for the volume of this trapezoid (rotates 360 degrees around the x-axis):

h= height

The area is equal to:
$\displaystyle A=\frac{1}{2}\cdot(r+R)\cdot h$

Volume of an object rotating around the x-axis:
$\displaystyle V_x=\pi\cdot\int_{a}^{b}f(x)^2\, dx$

Judging by the graph, it starts from x=0, and it ends on x=h, therefore the interval is [0;h].

The area function is the equation for a trapezoid, therefore, the function.

Conclusion:
$\displaystyle V_x=\pi\cdot\int_{0}^{h}(\frac{1}{2}\cdot(r+R)\cdo t h)^2\, dh$

Did I solve this correctly? I'm afraid, the "dh" is wrong, and generally just unsure if I solved this correctly.

2. Originally Posted by No Logic Sense
Question:
Find a formula for the volume of this trapezoid (rotates 360 degrees around the x-axis):

h= height

The area is equal to:
$\displaystyle A=\frac{1}{2}\cdot(r+R)\cdot h$

Volume of an object rotating around the x-axis:
$\displaystyle V_x=\pi\cdot\int_{a}^{b}f(x)^2\, dx$

Judging by the graph, it starts from x=0, and it ends on x=h, therefore the interval is [0;h].

The area function is the equation for a trapezoid, therefore, the function.

Conclusion:
$\displaystyle V_x=\pi\cdot\int_{0}^{h}(\frac{1}{2}\cdot(r+R)\cdo t h)^2\, dh$

Did I solve this correctly? I'm afraid, the "dh" is wrong, and generally just unsure if I solved this correctly.
Hi

You did not solve correctly !
$\displaystyle V_x=\pi\cdot\int_{a}^{b}f(x)^2\, dx$
This formula is applied to solid whose frontier is generated by the rotation of a curve whose equation is y=f(x) around x axis
The function to be taken into account is therefore the line joining the point (0,R) to the point (h,r)
You have to integrate
$\displaystyle V_x=\pi\cdot\int_{0}^{h}(R+\frac{r-R}{h}\cdot x)^2\, dx$

3. Ah, okay, I understand now. Thank you.

4. Hello, No Logic Sense!

Sorry, wrong set-up . . .

Find a formula for the volume of this trapezoid (rotates 360 degrees around the x-axis):

The slope of the slanted line is: $\displaystyle m = \tfrac{r-R}{h}$

The equation of the line is: .$\displaystyle y \:=\:\tfrac{r-R}{h}x + R$

We want to rotate the area under the line about the x-axis (from 0 to h).

The formula is: .$\displaystyle V \;=\;\pi\int^h_0\left(\tfrac{r-R}{h}x + R\right)^2\,dx$