evaluate : integral {-r; -r+d} sqrt (r^2 - y^2) dy
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The symbols r and d represent constants.
The symbol y is the variable.
Do you know how to determine the antiderivative of r^2 - y^2, with respect to y?
If so, then evaluate the definite integral in the usual way by replacing y with the expression d - r (as the upper bound) and with the expression -r (as the lower bound).
If you do not know how to determine the antiderivative of r^2 - y^2, with respect to y, then let me know.
You did not ask any questions. You did not make any statements about what you already know.
Therefore, I am not able to determine at what point in this exercise you're stuck.
The symbol y is the variable.
Do you know how to determine the antiderivative of r^2 - y^2, with respect to y?
If so, then evaluate the definite integral in the usual way by replacing y with the expression d - r (as the upper bound) and with the expression -r (as the lower bound).
If you do not know how to determine the antiderivative of r^2 - y^2, with respect to y, then let me know.
You did not ask any questions. You did not make any statements about what you already know.
Therefore, I am not able to determine at what point in this exercise you're stuck.
i was trying to solve it as a circle where r^2=X^2 + Y^2
then x= sqrt of (r^2 - y^2) which is the original integral
then i got integral of X dy
then i took u=x du = dx
dv=dy v=y
then applied
uv- integral vdu
is that right or just a mess
thank u again
then x= sqrt of (r^2 - y^2) which is the original integral
then i got integral of X dy
then i took u=x du = dx
dv=dy v=y
then applied
uv- integral vdu
is that right or just a mess
thank u again