What the area of the figure bounded by the given curve y=(x*(sqrt36-x^2)) (0_< x_< 6)?

Answer 1

Samantha Adkison

The area is #72# square units.

We have:

#A = int_0^6 xsqrt(36 - x^2)dx#

Now let #u = 36 - x^2#. Then #du = -2xdx# and #dx= (du)/(-2x)#.

#A = int_36^0 xsqrt(u) * (du)/(-2x)#

#A = 1/2int_0^36sqrt(u) du#

#A = 1/2[2/3u^(3/2)]_0^36#

#A = 1/2[2/3(36- x^2)^(3/2)]_0^6#

#A = (1/2)2/3(36 - 0)^(3/2)#

#A = (1/2)144#

#A = 72#

Hopefully this helps!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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What the area of the figure bounded by the given curve y=(x*(sqrt36-x^2)) (0_&lt; x_&lt; 6)?

What the area of the figure bounded by the given curve y=(x*(sqrt36-x^2)) (0_< x_< 6)?