Math  /  Geometry

QuestionUse geometry to evaluate the definite integral. 110121x2dx110121x2dx=\begin{array}{l} \int_{-11}^{0} \sqrt{121-x^{2}} d x \\ \int_{-11}^{0} \sqrt{121-x^{2}} d x= \end{array} \square (Simplify your answer. Type an exact answer, using π\pi as nee

Studdy Solution

STEP 1

What is this asking? We need to find the area under the curve y=121x2y = \sqrt{121 - x^2} from x=11x = -11 to x=0x = 0. Watch out! Remember that the equation of a circle is x2+y2=r2x^2 + y^2 = r^2, and we're only dealing with the positive square root, so we're looking at the top half of a circle!

STEP 2

1. Rewrite the equation.
2. Visualize the area.
3. Calculate the area.

STEP 3

Let's **rewrite** our equation y=121x2y = \sqrt{121 - x^2} to see what shape it represents.
If we square both sides, we get y2=121x2y^2 = 121 - x^2.

STEP 4

Now, let's **add** x2x^2 to both sides to get x2+y2=121x^2 + y^2 = 121.
This looks familiar!
It's the equation of a circle centered at the origin (0,0)(0, 0) with a radius of r=121=11r = \sqrt{121} = \mathbf{11}.
Since our original equation was y=121x2y = \sqrt{121 - x^2}, we're only considering the **positive** square root, meaning we're looking at the top half of the circle.

STEP 5

We want the area under this semi-circle from x=11x = \mathbf{-11} to x=0x = \mathbf{0}.
Imagine the circle.
From x=11x = -11 to x=0x = 0 is the left half of the semi-circle!

STEP 6

The area of a full circle is given by A=πr2A = \pi r^2.
Since we have a semi-circle (half a circle), its area is 12πr2\frac{1}{2} \pi r^2.
We're interested in the left half of this semi-circle, which is a quarter of the full circle!

STEP 7

So, the area we're looking for is 14\frac{1}{4} of the area of the full circle.
With our radius r=11r = \mathbf{11}, the area of the full circle is π112=121π\pi \cdot 11^2 = \mathbf{121\pi}.

STEP 8

Therefore, the area of the quarter-circle is 14121π=121π4\frac{1}{4} \cdot 121\pi = \mathbf{\frac{121\pi}{4}}.

STEP 9

The area under the curve y=121x2y = \sqrt{121 - x^2} from x=11x = -11 to x=0x = 0 is 121π4\frac{121\pi}{4}.

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