Maths Ext help (1 Viewer)

cossine

Active Member
So the integrand is area of the cross section. So just integrate that.

5uckerberg

Well-Known Member
So okay this is another find the volume of the cylinder type of question.

Step 1 Recognise which axis is the region rotated on.
Step 2 Know the volume of the cone and note that the x-axis represents the height using the concept of the cone through the rotation about the x-axis in $\bg_white y=x$
Step 3 Write in $\bg_white \pi\int_{0}^{1}y^{2}dx$ because the boundaries are 0 and 1 which is the height thus explaining this part $\bg_white \int_{0}^{1}$
Step 4 Write in $\bg_white \pi\int_{0}^{1}x^{2}\left(1-x^{2}\right)^{2}dx$
Step 5 Expand $\bg_white \left(1-x^{2}\right)^{2}=1-2x^{2}+x^{4}$
Step 6 Multiply by $\bg_white x^{2}$
Step 7 Integrate $\bg_white \pi\int_{0}^{1}x^{2}-2x^{4}+x^{6}dx$
Step 8 Solve $\bg_white \frac{1}{3}-\frac{2}{5}+\frac{1}{7}$ by completing the calculation.

What you will see here @Farhanthestudent005 is that you need to know what the volume of the cylinder is but with a minuscule height because at the end of the day when you rotate a graph along either the x or y-axis or z-axis if you are in 2nd-year uni or above you are inevitably going to have circles and the area of the circle is $\bg_white \pi{r^{2}}$ so then to find the volume of the cylinder you have to have $\bg_white \pi{r^{2}}h$ or to recognise the pattern it will be $\bg_white \pi\int_{lower bound}^{upper bound}\left(x \cup y\right)^{2}d\left(Rotation along the ... axis\right)$

5uckerberg

Well-Known Member
If you have any difficulties feel free to reply to the post and someone will help you along your mathematical journey.