Question

What is the area enclosed by `r=-sin(3theta-(7pi)/4)` between `theta in [pi/8,(pi)/4]`?

Answer 1

`1/4(pi/8-1/(6sqrt2))=0.0687` areal units, nearly. A part of this area cannot be seen in Socratic graph. r-negative loop in `Q_1` contributes this part. See graph for clarity.

Explanation:

As depicted by the reliable Socratic graph, the area is in #Q_1

and Q_3 loops of this limacon. r-negative loops are not included in

this utility. This deserves my appreciation. Yet, conventional r-

negative loops are included in some other graphic devices.

In my opinion, `theta = pi/8 and pi/4` for the limits, limit the

boundary lines to half lines only, in the respective directions. So, the

area under reference is in the first quadrant only. I include the

unseen r-negative `Q_1` loop also.

Area = 1/2`int r^2 d theta`, with r for the limacon and

`theta` from `pi/8` to `pi/4`

`=1/2int sin^2(3theta-7/4pi) d theta`, for the limits

`=1/4 int (1-cos(6theta-7/2pi) d theta`, for the limits

`1/4[theta-1/6sin(6theta-7/2pi)]`, between the limits

`=1/4[pi/8-1/6(0-sin(-11/4pi)]`

`=1/4(pi/8-1/(6sqrt2))`

`=0.0687` areal units, nearly.

If the opposite area from two ( r-positive and r-negative ) loops in

`Q_3` ) are also included, it is double this area.

Note : My hands were unable to move cursor on my computer for

editing my answer, owing to editing my answer by another. I had

expressed my displeasure over this, more than once. I request

others to give comments or another answer, instead of causing

delay to my service to Socratic. I am eager to see good answers

from others. I am sparing my very precious time of my last and great

quarter of my life, here. Please, do note make me stop this.

graph{((x^2+y^2)^2.5+0.707(x^4+4x^3y-6x^2y^2-4xy^3+y^4))(y-x)(y-0.4142x)=0 [-2, 2, -1, 1]}