Calculus I Problem , Need Help

Question

Let $~R~$ be the finite region in the first quadrant bounded by the graphs of $~y = 4x + 1~$ and $~y = x^2 + 1~$ and below the horizontal line $~y = 5~$. Set up, but do not evaluate, definite integrals which represent the given quantities. Use proper notation.

$a)~~$ The volume of the solid with base $~R~$ whose cross-sections perpendicular to the $~x$-axis are isosceles right triangles, with the hypotenuse lying in the $~xy$-plane.

I'm not sure what this question is asking, can someone please help me to answer this question?

Answer 1

Here is a graph of $R$ and its boundary functions.

The solid has this region as its base. Suppose we take cross-sections of this solid that are perpendicular to the $xy$-plane (that is, perpendicular to the plane of the above graph), and also perpendicular to the $x$-axis. For perspective, such cross-sections would look like vertical segments in the above graph within $R$.

These vertical segments through $R$ are hypotenuses of isosceles right triangles (each right triangle is a full cross-section of the solid). Imagine the legs of these right triangles as protruding out of the plane of $R$ (the graph on your computer monitor) so that the vertices of the right angles are outside your computer monitor (say, between the monitor and where you are sitting). Visualizing these cross-sections is important.

Cavalieri's principle tells us that if we simply add up the areas of these cross-sections, we will obtain the volume of this solid. To compute this, we split $R$ into two sections, $R_1$ over $0<x<1$ and $R_2$ over $1<x<2$.

The legs of an isosceles right triangle with hypotenuse $h$ have length $\frac{h}{\sqrt{2}}$. So the area of an isosceles right triangle with hypotenuse $h$ is $\frac{1}{2}\left(\frac{h}{\sqrt{2}}\right)^2=\frac{h^2}{4}$.

The vertical segments (hypotenuses) in $R_1$ for a given $x$ are equal to $4x+1-(x^2+1)=4x-x^2$. So for a given $x$, the area of the cross-section is given by $\frac{1}{4}(4x-x^2)^2$. Summing these up continuously over the interval bounding $R_1$, we find the volume of one portion of the solid to be $$\frac{1}{4}\int_{0}^{1}{(4x-x^2)^2\textrm{ d}x}.$$ Similarly, you can find another definite integral that gives the volume of the the portion of the solid which has a base of $R_2$. Adding those two volumes will give you the volume of the entire solid.

Answer 2

You have to find the volume of an object whose base is the area bounded by $y=x^2+1$, $y=5$, and $y=1+4x$

The cross area section area $A(x)$ is the area of an isosceles right triangle with the hypotenuse being the vertical line segment bounded by the base lines.

The volume is $$ V= \int _0^2 A(x) dx$$

The Region is divided into two regions and the you have to evaluate two integrals as well.

$$V= \int _0^1 (1/4)(4x-x^2)^2 dx +\int _1^2 (1/4)(4-x^2)^2 dx$$

You should know how to evaluate the integrals.